Is NCERT enough for Class 12 Physics?

Yes, NCERT is more than enough for Class 12 Physics. The students can practice the CBSE Class 12 Physics Important Questions given on this page.

Where can I find CBSE Class 12 Physics Important Questions?

You can find the CBSE Class 12 Physics Important Questions here. We have given the CBSE Class 12 Physics Important Questions based on the latest CBSE Exam pattern on this page.

When will CBSE conduct CBSE Class 12 Physics Exam 2024?

CBSE Class 12 Exam Physics exam will be held on 4 March 2024.

CBSE Physics Class 12 Important Questions and State Board with Answers PDF

Table of Contents

The Physics class 12 important questions given below in the article will prove to be useful for students who are going to take the CSBE board exam 2024 as well as for the other board exams. The CBSE Physics Class 12 Important Questions pose a great challenge to students if they do not practice it well. For the convenience of students, below we have provided the

Physics Class 12 Important Questions

As Central Board of Secondary Education has announced the CBSE Exam Date 2023. The students have to practice Class 12 Physics Important Questions given on this page to get good marks in the exam. On this page, we have given Class 12 Physics Important Questions as per the latest exam pattern of CBSE. We have segregated 2 marks, 3 marks, and 5 marks questions from each chapter of Class 12 Physics and given them on this page to help students in their Class 12 Exam preparation. The students appearing in CBSE Class 12 exam can bookmark this page to get the latest updates from the Central Board of Secondary Education.

Physics Class 12 Important Questions CBSE & State Board with Answers

Preparing for any exam requires familiarity with the exam pattern. If required, students will be provided a log table, and the use of calculators in exams will be restricted. The CBSE Term 2 Physics question paper is divided into three sections: Section A, Section B, and Section C. Check the details given below:

Section A: In this section, each question carries 2 marks.

Section B: In this section, each question carries 3 marks.

Section C: In this section, each question carries 5 marks.

Physics Class 12 Important Questions CBSE & State Board with Answers – 2 Marks

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Electromagnetic Waves

Q. The charging current for a capacitor is $0.25 Ω$ . What is the displacement current across its plates?

Ans: Displacement current is equivalent to and is the same as charging current $025 A$ .

Q. Write the following radiations in a descending order of frequencies: red light, x-rays, microwaves, radio waves

Ans: The descending order of frequencies: red light, x-rays, microwaves, radio waves are given as: X-rays > Red light >Microwaves >Radio Waves

Q. How does the frequency of a beam of ultraviolet light change, when it goes from air into glass?

Ans: We know that the frequency of UV light is unaffected by the frequency of a beam of ultraviolet light change and when it goes from air into glass.

Q. What is the ratio of speed of gamma rays and radio waves in vacuum?

Ans: As we know, the ratio of speed of gamma rays and radio waves in vacuum is one

Q. It is necessary to use satellites for long distance TV transmission. Why?

Ans: Because television signals are not reflected by the ionosphere layer of the atmosphere, TV broadcasts from air earth stations must be reflected back to the ground via an artificial satellite.

Q. What is the role of ozone layer in the atmosphere?

Ans: The main role of the ozone layer in the atmosphere is that it absorbs all harmful UV rays and shields us from their detrimental effects.

Q. What is the nature of waves used in radar?

Ans: The characteristics of the waves utilized in radar employs microwaves.

Q. What physical quantity is the same for $X$ rays of wavelength $10 - 10 m$ , red light of wavelength 6800 A and radio waves of wavelength $500 m$ ?

Ans: As we know that the speed of light $(3 \times 10^{8} m / s)$ all wavelengths are the same in a vacuum. In the vacuum and they are unaffected by the wavelength.

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Optics

Q. How do the focal lengths of a lens change with increase in the wavelength of the light?

Ans: We know that

1 f = (μ - 1) (1 R 1 - 1 R 2)

Also,

μ \propto 1 λ

Clearly, when wavelength $(λ)$ increases, refractive index

(μ)

decreases.

Similarly, $μ \propto f$ .

Clearly, as refractive index

(μ)

decreases, focal length $(f)$ increases.

Q. Show with a ray diagram, how an image is produced in a total reflecting prism?

Ans: Consider the two rays from the object PQ, as shown below.

They undergo total internal reflection firstly at the face AB and then at BC forming the final image

P' Q'

(real and inverted image).

Q. The radii of the curvature of the two spherical surfaces which is a lens of required focal length are not the same. It forms an image of an object. The surfaces of the lens facing the object and the image are interchanged. Will the position of the image change?

Ans: As we know that,

1 f = (μ - 1) (1 R 1 - 1 R 2)

When the radii of curvature

R 1

and

R 2

are interchanged, the focal length of the lens also changes. Hence, the position of the image will reduce gradually.

Q. A thin converging lens has focal length (f) when illuminated by violet light. State with reason how the focal length of the lens will change if violet light is replaced by red light.

Ans: We know that,

1 f = (n - 1) (1 R 1 - 1 R 2)

Since, $n$ for violet is more that $n$ for red colour, and since $m \propto f$ , we can say that the focal length of the lens will decrease when violet light is replaced by red light.

Q. What is the shape of the wave front when light is diverging from a point source?

Ans: When light diverges from a point source the wave front would show spherical shape.

Q. State the conditions that must be satisfied for two light sources to be coherent.

Ans: The conditions to be satisfied are the following:

a) They must emit waves continuously of same wavelengths.

b) The phase difference between the waves must be zero or constant.

Q. You read a newspaper, because of the light it reflects. Then why do you not see even a faint image of yourself in the newspaper?

Ans: We know that image is formed due to the regular reflection of light.

However, when we read a newspaper, there is diffused (irregular) reflection of light, thus we are not able to see even a faint image of ourselves on the newspaper.

Q. Let us list some of the factors, which could possibly influence the speed of wave propagation:

Nature of the source
Direction of propagation
Motion of the source and/or observer
Wave length
Intensity of the wave. On which of these factors, if any, does
The speed of light in vacuum

Ans: We know that the speed of light in vacuum, $3 \times 10^{8} m / s$ (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. So, the given factor does not affect the speed of light in a vacuum.

Q. The speed of light in a medium (say, glass or water), depends.

Ans: Out of the listed factors, the speed of light in a medium would depend on the wavelength of light in that medium.

Q. Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?

Ans: Myopia and hypermetropia are common eye defects.

A myopic or hypermetropic person need not necessarily suffer a partial loss in their eyes’ ability of accommodation.

Myopia occurs when the eye-balls engage in elongation from the front to the back whereas hypermetropia occurs when the eye-balls shorten themselves.

On the other hand, when the eye-lens completely loses its ability of adjusting itself, then the defect is called presbyopia.

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Dual Nature of Radiation and Matter

Q. Mention one physical process for the release of electrons from the surface of a metal.

Ans: Photoelectric emission.

The phenomenon in which the electrons from the surface of a metal are given energy in form of electromagnetic waves and they are ejected out, this phenomenon is called the photoelectric emission.

Q. The maximum kinetic energy of photoelectron is?

2.8 e V

Q. What is the value of stopping potential?

Ans: Given an electron that is moving with a kinetic energy. For it to be not ejected, it has to be held back using a stopping potential

V 0

The relation between the two is:

$K E = e V_{o} = 2.8 e V$

$\Rightarrow V_{o} = 2.8 V$

Q. Ultraviolet light is incident on two photosensitive materials having work function $ϕ_{1}$ and $ϕ_{2}$ ( $ϕ_{1} > ϕ_{2}$ ). In which of the case will K.E. of emitted electrons be greater? Why?

Ans: According to the energy balance equation of the photoelectric effect $h v = ϕ_{o} + K . E$

If $ϕ_{1} > ϕ_{2}$ thus K.E. will be more for second surface whose work function is less.
Q. Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of incident radiations.

Ans: Suppose

$ν_{o}$ is the threshold of frequency or cut off frequency;

$V_{o}$ is the corresponding stopping potential

Q. How does the stopping potential applied to a photocell change if the distance between the light source and the cathode of the cell is doubled?

Ans: Intensity of light drops quadratically with distance. However, the stopping potential does not depend on the intensity of the light. Hence it is independent of distance as well.

Q. On what factor does the retarding potential of a photocell depend?

Ans: The retarding photocell depends upon the frequency of the incident light

Q. Electron and proton are moving with same speed, which will have more wavelength?

Ans: Since the wavelength is inversely proportional to the square root of the mass of the body, $λ \propto \frac{1}{\sqrt{m}}$ . So, electrons being lighter will have more wavelengths.

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Atoms and Nuclei

Q. Complete the following nuclear reactions:

a) 4Be9+1H1→3Li6+……….

Ans: 4Be9+1H1→3Li6+2He4

b) 5B10+2He4→7N13+……….

Ans: 5B10+2He4→7N13+0n1

Q. What is the Q-value of a nuclear reaction?

Ans: Q−value=(Mass of reactants−Mass of products)

Q. The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 9546∘A, 6463∘A and 1216∘A. Which one of these wavelengths belongs to the Lyman series?

Ans: 1216∘A belong to the Lyman series.

Q. Name the series of hydrogen spectrum lying in ultraviolet and visible region.

Ans: Lyman series lies in ultraviolet region while Balmer series lies in visible region.

Q. State the limitations of Bohr’s atomic model.

Ans: The limitations of Bohr’s atomic model are:

(1) It does not give any indication regarding the arrangement and distribution of electrons in an atom.

(2) It could not account for the wave nature of electrons.

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Electronic Devices

Q. What type of impurity is added to obtain n-type semiconductor?

Ans: Pentavalent atoms (group – $15$ ) like Phosphorus (P), Arsenic (As), etc.

Q. Doping of silicon with indium leads to which type of semiconductor?

Ans: Doping of Silicon with Indium produces a p-type semiconductor as Indium is a trivalent impurity.

Q. How does the energy gap of an intrinsic semiconductor vary, when doped with a trivalent impurity?

Ans: An acceptor energy level is formed in the forbidden energy gap above the valence band when an intrinsic semiconductor is doped with a trivalent impurity. Due to this, electrons quickly jump to the acceptor energy level.

Q. How does the width of the depletion layer of p-n-junction diode change with decrease in reverse bias?

Ans: The width of the depletion layer will decrease with decrease in reverse bias.

Q. Under what condition does a junction diode work as an open switch?

Ans: A junction diode works as an open switch when it is connected under reverse bias conditions.

Q. Which type of biasing gives a semiconductor diode very high resistance?

Ans: Reverse biasing gives a semiconductor diode very high resistance.

Q. Define current amplification factor in a common – emitter mode of transistor.

Ans: Current amplification factor is the fraction of a small change in collector current to the slight difference in base current at constant collector-emitter junction voltage.

Q. Why is a semiconductor damaged by a strong current?

Ans: When a strong current traverses through a semiconductor, a large amount of heat is generated, which breaks the covalent links in the semiconductor. This causes the semiconductor to get damaged.’

Physics Class 12 Important Questions CBSE & State Board with Answers

Physics Class 12 Important Questions CBSE & State Board with Answers– 3 Marks

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Electromagnetic Waves

Q. Write the characteristics of emf waves? Write the expression for velocity of electromagnetic waves in terms of permittivity and permeability of the medium?

Ans: The Characteristics of emf waves are as follows :

(1) It travels in free space with speed of light $c = 3 \times 10^{6} m / s$ ,

(2) Electromagnetic waves are transverse in nature.

Velocity of emf waves in vacuum C $= \frac{1}{\sqrt{μ_{0} \in_{6}}}$

Q. The electric field of a plane electromagnetic wave in vacuum is represented $b y$ –

$E x = 0, E y = 0.5 \cos (2 π \times 10^{8} (t - x / c))$ and $E_{z} = 0$

a) What is the direction of propagation of electromagnetic wave?

Ans: For the given equation It represents the wave is propagating along $+ \bar{x}$ – axis

b) Determine the wavelength of the wave?

Ans: Now, by Comparing equation with the standard one

$Ey = E_{0} \cos w (t - x / c)$

$w = 2 π \times 10^{2}$

$z^{'} \vec{t} t = 2^{'} z \times 10^{8}$

$v = 10^{8}$

$\Rightarrow λ = \frac{c}{v} = \frac{3 \times 10^{8}}{10^{8}}$

$λ = 3 m$

c) Compute the component of associated magnetic field

Ans: As it is associated magnetic field to electric field and the direction of propagation. Since wave is propagating along $x =$ axis, electric field is along, $y =$ axis

Thus, magnetic field is along $z =$ axis

$\Rightarrow B_{x} = 0, B_{y} = 0$

$B_{z} = B_{0} \cos (2 π \times 10^{1} (t - x / x))$

$B_{z} = \frac{E_{0}}{C} \cos (2 π \times 10^{1} (t - x / x))$

Q. Find the wavelength of electromagnetic waves of frequency $5 \times 10^{18} H z^{in free space .}$ Give its two applications.

Ans: By Using $C = λ v$

$v = 5 \times 10^{15} Hz$

$λ = \frac{3 \times 10^{1}}{5 \times 10^{^{\circ} 6}}$

$λ = 0.6 \times 10^{- 11} = 6 \times 10^{- 12} m$

These are Gamma Rays.

Applications for gamma rays :

(1) These rays are used to get information regarding atomic structure.

(2) They have very high penetrating power so they are used for detection purpose

Q. State the condition under which a microwave oven heats up food items containing water molecules most efficiently?

Ans: (1) Microwaves must have a frequency that is equal to the resonant frequency of the water molecules in the food item.

(2) Name the radiations which are next to these radiations in emf. Spectrum having (a) Shorter wavelength (b) Longer wavelength

(a) visible light is the radiation are the radiation next to shorter wavelength.

(b) Microwaves light is the radiation are the radiation next to longer wavelength.

Q. Long distance radio broadcasts use short-wave bands. Why?

Ans: Shortwave bands are used for long-distance radio broadcasts because only these bands can be refracted by the ionosphere.

Q. It is necessary to use satellites for long distance TV transmission. Why?

Ans: Because television signals have high frequencies and energy, satellites are required for long-distance TV transmissions. As a result, the ionosphere does not reflect these signals. As a result, satellites aid in the reflection of television transmissions.

Q. Optical and radio telescopes are built on the ground but -ray astronomy is possible only from satellites orbiting the earth. Why?

Ans: As far as -ray astronomy is concerned, -rays are absorbed by the atmosphere. Visible and radio waves, on the other hand, can pass through it. As a result, optical and radio telescopes are constructed on the ground, but X-ray astronomy is only conceivable thanks to satellites orbiting the Earth.

Q. The small ozone layer on top of the stratosphere is crucial for human survival. Why?

Ans: The ozone layer at the top of the atmosphere is critical for human survival because it absorbs damaging ultraviolet radiation from the sun and keeps it from reaching the Earth’s surface.

Q. If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

Ans: There would be no greenhouse effect on the Earth’s surface if it did not have an atmosphere. As a result, the Earth’s temperature would swiftly drop, making it cold and uncomfortable to live on and difficult for human survival.

Q. Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

Ans: On the surface of the Earth, a global nuclear war would be catastrophic. The Earth will experience a harsh winter following a nuclear war because the war will produce clouds of smoke that will cover the majority of the sky, preventing solar light from reaching the atmosphere. It will also contribute to the ozone layer’s depletion.

Physics Class 12 Important Questions For UP Board with Answers Unit – Optics

Q. Let us list some of the factors, which could possibly influence the speed of wave propagation:

Nature of the source
Direction of propagation
Motion of the source and/or observer
Wave length
Intensity of the wave. On which of these factors, if any, does

The speed of light in vacuum

Ans: We know that the speed of light in vacuum, $3 \times 10^{8} m / s$ (approximately) is a universal constant. It is not affected by the motion of the source, the observer, or both. So, the given factor does not affect the speed of light in a vacuum.

The speed of light in a medium (say, glass or water), depends.

Ans: Out of the listed factors, the speed of light in a medium would depend on the wavelength of light in that medium.

Q. For sound waves, the Doppler formula for frequency shift differs slightly between the two situations given and explain why this should be so. Would you expect the formulas to be strictly identical for the two situations in case of light travelling in a medium?

source at rest; observer moving, and
source moving; observer at rest. The exact Doppler formulas for the case of light waves in vacuum are, however, strictly identical for these situations.

Ans: We know that sound waves can propagate only through a medium. The two given situations are not scientifically identical as the motion of an observer relative to a medium is different in the given two situations. So, the Doppler formulas for the given two situations cannot be the same.

Now, in case of light waves, sound can travel in a vacuum. But in vacuum, the above two cases are found to be identical as the speed of light is independent of the motion of the observer and the motion of the source. While, when light travels in a medium, the above two cases are not identical as the speed of light would now depend on the wavelength of the medium.

Q. In double-slit experiment using light of wavelength $600 n m$ , the angular width of a fringe formed on a distant screen is $0.1^{\circ}$ . What is the spacing between the two slits?

Ans: We are given:

Wavelength of light used, $λ = 6000 nm = 600 \times 10^{- 9} m$

Angular width of fringe, $θ = 0.1^{\circ} = 0.1 \times \frac{λ}{180} = \frac{3.14}{1800} rad$

Angular width of a fringe is related to slit spacing $(d)$ as:

θ = λ d

\Rightarrow d = λ θ

∴ d = 600 \times 10 - 9 3.14 1800 = 3.44 \times 10 - 4 m

Therefore, the spacing between the slits is found to be $3.44 \times 10^{- 4} m$ .

Q. In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angles of $n λ / a$ . Justify this by suitably dividing the slit to bring out the cancellation.

Ans: Consider that a single slit of width $d$ is divided into $n$ smaller slits.

That is, width of each slit, $d = \frac{d}{n}$

Angle of diffraction is given by the relation,

θ = d d λ d = λ d

Now, each of these infinitesimally small slits send zero intensity in direction $θ$ . Thus, the combination of these slits will give zero intensity.

Q. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?

Ans: In the given case, the person is able to see vertical lines more distinctly than horizontal lines.

This means that the refracting system (cornea and eye-lens) of the eye is not working effectively in different planes. This defect is called astigmatism.

The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient.

Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

Q. Answer the following questions:

When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TV screen. Suggest a possible explanation.

Ans: We know that weak radar signals are sent by a low flying aircraft and this can interfere with the TV signals received by the antenna. As a result of this, the TV signals might get distorted. Hence, when a low flying aircraft passes overhead, we could sometimes notice a slight shaking of the picture on our TV screen.

As you have learnt in the text, the principle of linear superposition of wave displacement is basic to understanding intensity distributions in diffraction and interference patterns. What is the justification of this principle?

Ans: For our understanding of intensity distributions and interference patterns, the principle of linear superposition of wave displacement is essential. This is because superposition follows from the linear character of a differential equation that is known to govern wave motion. Let $y_{1}$ and $y_{2}$ are the solutions of the second order wave equation, then any linear combination of $y_{1}$ and $y_{2}$ might also be the solution of the wave equation.

Q. What is polarization of light? What type of waves show the property of polarization? Name any two methods to produce plane polarized light.

Ans: The phenomenon by virtue of which the vibrations of a light vector is restricted in a particular direction in a plane perpendicular to the direction of propagation of light is called polarisation of light. Transverse waves are known to show the property of polarisation. Two methods to produce plane polarised light are:

Polarisation by Reflection and
Polarization by scattering.

Q. Draw the curve depicting variation of intensity in the interference pattern in young’s double slit experiment. State conditions for obtaining sustained interference of light?

Ans: The curve could be drawn as,

Conditions for sustained interference of light are:

Two sources causing the interference must be coherent sources of light.
Two sources causing the interference should have nearly equal amplitudes and intensities and should be monochromatic.

Q. What is the shape of the wave front in each of the following cases:

Light diverging from a point source.

Ans: The shape of the wave front in case of a light that is diverging from a point source is spherical. The wave front emanating from a point source would be as shown in the below figure.

(Image will be uploaded soon)

Light emerging out of a convex lens when a point source is placed at its focus.

Ans: The shape of the wave front in case of a light emerging out of a convex lens when a point source is placed at its focus would be a parallel grid. This would be as shown in the below figure.

The portion of the wave front of light from a distant star intercepted by the Earth.

Ans: The portion of the wavefront of light from a distant star intercepted by the Earth would be a plane.

Class 12 Physics Important Questions For Board Exam 2024 with Answers Unit – Dual Nature of Radiation and Matter

Q. Show on a plot the nature of variation of photoelectric current with the intensity of radiation incident on a photosensitive surface.
Answer:

Q. Figure shows a plot of 1V√, where V is the accelerating potential, vs. the de-Broglie wavelength ‘λ’ in the case of two particles having same charge ‘q’ but different masses m₁ and m₂. Which line (A or B) represents a particle of larger mass?

Answer:
B line represents particle of larger mass because slope ∝1m√.

Q. Find the ratio of de-Broglie wavelengths associated with two electrons accelerated through 25 V and 36 V.
Answer:

Q. Define intensity of radiation on the basis of photon picture of light. Write its S.I. unit.
Answer:
It is the number of photo-electrons emitted per second per unit area.
SI unit : m^-2S^-1

Q. The graph shows the variation of stopping potential with frequency of incident radiation for two photosensitive metals A and B. Which one of the two has higher value of work- function? Justify your answer.

Answer:
Metal ‘A’, because of higher threshold frequency for it.

Q. The graph shows variation of stopping potential V₀ versus frequency of incident radiation v for two photosensitive metals A and B. Which of the two metals has higher threshold frequency and why?

Answer:
Metal ‘A’, because of higher threshold frequency for it.

Q. An electron is revolving around the nucleus with a constant speed of 2.2 × 10⁸ m/s. Find the de-Broglie wavelength associated with it
Answer:

Q. Draw a plot showing the variation of de Broglie wavelength of electron as a function of its K.E.
Answer:

Q. Name the phenomenon which shows the quantum nature of electromagnetic radiation.

Answer:
Photoelectric Effect is the phenomenon which shows the quantum nature of electro-magnetic radiation.

Q. State one factor which determines the intensity of light in the photon picture of light.
Answer:
The factor determining the intensity of light is number of electrons emitted per second.

Q. State one reason to explain why wave theory of light does not support photoelectric effect. (Comptt. Delhi 2014)
Answer:
One reason why wave theory of light does not support photoelectric effect is that the kinetic energy of photo electrons does not depend on the intensity of incident light.

Q. Write three basic properties of photons which are used to obtain Einstein’s photoelectric equation. Use this equation to draw a plot of maximum kinetic energy of the electrons emitted versus the frequency of incident radiation.

Answer:
Properties.
Einstein’s photoelectric equation is K_max = hv – ϕ₀
(i) We find K_max depends linearly on V only. It is independent of intensity of radiation.

(iii) Greater the number of energy quanta, greater is the number of photoelectrons. So, photoelectric current is proportional to intensity.

Plot of maximum kinetic energy vs. frequency

Q. (i) Define the term ‘threshold frequency’ as used in photoelectric effect.
(ii) Plot a graph showing the variation of photoelectric current as a function of anode potential for two light beams having the same frequency but different intensities I₁ and I₂ (I₁ > I₂).
Answer:
(i) Threshold frequency. The minimum frequency v₀ which the incident light must possess so as to eject photoelectrons from a metal surface, is called threshold frequency of the metal.

Q. A proton and an a-particle have the same de- roglie wavelength. Determine the ratio of
(i) their accelerating potentials
(ii) their speeds.
Answer:
The de-Broglie wavelength for a proton is,

Q. Using the graph shown in the figure for stopping potential v/s the incident frequency of photons, calculate Planck’s constant.

Answer:

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Atoms and Nuclei

Q. (a) The mass of a nucleus in its ground state is always less than the total mass of its constituents – neutrons and protons. Explain.
(b) Plot a graph showing the variation of potential energy of a pair of nucleons as a function of their separation.
Answer:
(a) When nucleons approach each other to form a nucleus, they strongly attract each other. Their potential energy decreases and becomes negative. It is this potential energy which holds the nucleons together in the nucleus. The decrease in’ potential energy results in the decrease in the mass of the nucleons inside the nucleus.

Q. A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5 MeV per nucleon. Calculate the energy Q released per fission in MeV.
Answer:

∴ Gain in binding energy for nucleon = 8.5 – 7.6 = 0.9 MeV
Hence total gain in binding energy per nucleus fission = 240 × 0.9 = 216 MeV

Q. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.
Answer:
Two important conclusions :
(i) Nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This explains constancy of the binding energy per nucleon for large-size nucleus.

(ii) Graph explains that force is attractive for distances larger than 0.8 fin and repulsive for distances less than 0.8 fm.

Q. Draw a plot of the binding energy per nucleon as a function of mass number for a large number of nuclei, 2 ≤ A ≤ 240. How do you explain the constancy of binding energy per nucleon in the range 30 < A < 170 using the property that nuclear force is short-ranged?
Answer:
(a) The constancy of the binding energy in the range 30 < A < 170 is a consequence of the fact that the nuclear force is short ranged.
If a nucleon can have a maximum of p neighbours within the range of nuclear force, its binding energy would be proportional to p. Since most of the nucleons in a large nucleus reside inside it and not on the surface, the change in binding energy per nucleon would be small. The binding energy per nucleon is a constant and is approximately equal to pk. The property that a given nucleon influences only nucleons close to it, is referred to as saturation property of the nuclear force.

(b) Nuclear force is short-ranged for a sufficiently large nucleus. A nucleon is under the influence of only some of its neighbours, which come within the range of the nuclear force. If a nucleon can have maximum of P neighbours within the range of nuclear force, its binding energy would be proportional to ‘P’ Thus on increasing ‘A’ by adding nucleons binding energy will remain constant.

Q. Using the curve for the binding energy per nucleon as a function of mass number A, state clearly how the release of energy in the processes of nuclear fission and nuclear fusion can be explained.
Answer:
1. Nuclear fission : Binding energy per nucleon is smaller for heavier nuclei than the middle ones i.e. heavier nuclei are less stable. When a heavier nucleus splits into the lighter nuclei, the B.E./nucleon changes (increases) from about 7.6 MeV to 8.4 MeV. Greater binding energy of the product nuclei results in the liberation of energy. This is what happens in nuclear fission which is the basis of the atom bomb.

2. Nuclear fusion : The binding energy per nucleon is small for light nuclei, i.e., they are less stable. So when two light nuclei combine to form a heavier nucleus, the higher binding energy per nucleon of the latter results in the release of energy.

Q. Complete the following nuclear reactions :

Answer:

Q. If both the number of protons and neutrons in a nuclear reaction is conserved, in what way is mass converted into energy (or vice verse)? Explain giving one example.
Answer:
Explanation for release of energy in a nuclear reaction : Since proton number and neutron number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of the nuclear reaction.
But total binding energy of nuclei on the left side need not be the same as that on the right hand side. The difference in binding energy causes a release of energy in the reaction.
Examples :

Q.

Answer:

Q. Define ionization energy.
How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge?
Answer:
Definition of ionization energy : “The minimum energy, required to free the electron from the ground state of the hydrogen atom, is known as Ionization Energy.”
The ionization energy is given by :

∴Ionization Energy will become 200 times,
∵ the mass of given particle is 200 times.

Q. Calculate the shortest wavelength of the spectral lines emitted in Balmer series.
[Given Rydberg constant, R = 10⁷ m^-1]

Answer:

Q. The electron, in a hydrogen atom, is in its second excited state.
Calculate the wavelength of the lines in the Lyman series, that can be emitted through the permissible transitions of this electron.
(Given the value of Rydberg constant, R = 1.1 × 107 m^-1)
Answer:

Question 23.
An α-particle moving with initial kinetic energy K towards a nucleus of atomic number z approaches a distance ‘d’ at which it reverses its direction. Obtain the expression for the distance of closest approach ‘d’ in terms of the kinetic energy of α-particle K. (Comptt. All India 2016)
Answer:

Question 24.
Find the ratio between the wavelengths of the ‘most energetic’ spectral lines in the Balmer and Paschen series of the hydrogen spectrum. (Comptt. All India 2016)
Answer:

Q.Define the distance of closest approach. An a-particle of kinetic energy ‘K’ is bombarded on a thin gold foil. The distance of the closest approach is V. What will be the distance of closest approach for an a-particle of double the kinetic energy?
Answer:
The distance of closest approach is defined as “the distance of charged particle from the centre of the nucleus, at which the whole of the initial kinetic energy of the (far off) charged particle gets converted into the electric potential energy of the system”.
Distance of closest approach (r_c) is given by

Q. Write two important limitations of Rutherford nuclear model of the atom.
Answer:
Important limitations of Rutherford Model :

According to Rutherford model, electron orbiting around the nucleus, continuously radiates energy due to the acceleration; hence the atom will not remain stable.
As electron spirals inwards; its angular velocity and frequency change continuously; therefore it. will emit a continuous spectrum.

Question 27.
Find out the wavelength of the electron orbiting in the ground state of hydrogen atom. (Delhi 2016)
Answer:

Q.
Find the wavelength of the electron orbiting in the first excited state in hydrogen atom. (Delhi 2016)
Answer:

Q. A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted.
Answer:

Q. The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 A Calculate the short wavelength limit for Balmer series of the hydrogen spectrum.

Answer:

12th Physics Important Questions Chapter Wise With Answers – 5 Marks

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Electromagnetic Waves

Q. In a plane electromagnetic wave, the electric field oscillates sinusoid ally at a frequency of $2.0 \times 10^{10} H z$ and amplitude $48 v_{m}^{- 1}$ .

(a) What is the wavelength of the wave?

(b) What is the amplitude of the oscillating magnetic field?

(c) Show that the average energy density of the E field equals the average energy density of the B field. (c $= 3 \times 10^{8} m s^{- 1}$ . )

Ans: As it is given that,

Frequency of the electromagnetic wave, $v = 2.0 \times 10^{10} Hz$

Electric field amplitude, $E_{0} = 48 V m^{- 1}$

Speed of light, $c = 3 \times 10^{3} m / s$

(a) What is the wavelength of the wave?

Ans: Now, Wavelength of a wave is given as:

$λ = \frac{c}{v}$

$= \frac{3 \times 10^{8}}{2 \times 10^{12}} = 0.015 m$

(b) What is the amplitude of the oscillating magnetic field?

Ans: Where,

Magnetic field strength is given as:

$B_{0} = \frac{E_{0}}{c}$

$= \frac{48}{3 \times 10^{2}} = 1.6 \times 10^{- 7} T$

(c) Show that the average energy density of the E field equals the average energy density of the B field. (c $= 3 \times 10^{8} m s^{- 1}$ . )

Ans: Then, Energy density of the electric field is given as:

$U_{E} = \frac{1}{2} \in_{0} E^{2} \dots (1)$

And, energy density of the magnetic field is given as:

$U_{B} = \frac{1}{2 μ_{0}} B^{2} \dots (2)$

Where,

ε 0

– Permittivity of free space

$μ_{0} =$ Permeability of free space

We have the relation connecting $E$ and $B$ as:

$E = c B \dots (3)$

Where,

$c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$

Now, Substituting equation (3) in equation (1), we get

Squaring both sides, we get

$\Rightarrow U_{E} = U_{B}$

Q. Suppose that the electric field amplitude of an electromagnetic wave is $E_{0} = 120 N / C$ and that its frequency is $v = 50.0 MHz$ . (a) Determine, $B_{0}, ω, k$ , and $λ$ . (b) Find expressions for $E$ and $B$ .

Ans: Given that ,

Electric field amplitude, $E_{0} = 120 N / C$

Frequency of source, $v = 50.0 MHz = 50 \times 10^{5} Hz$

Speed of light, $c = 3 \times 10^{8} m / s$

(a) Magnitude of magnetic field strength is given as:

$B_{0} \frac{E_{0}}{c}$

$= \frac{120}{3 \times 10^{2}}$

$= 4 \times 10^{- 7} T = 400 n T$

Angular frequency of source is given as:

$ω = 2 m = 2 π \times 50 \times 10^{6}$

$= 3.14 \times 108 rad / s$

Propagation constant is given as:

$k = \frac{(i)}{c}$

$= \frac{3.14 \times 10^{8}}{3 \times 10^{8}} = 1.05 rad / m$

Now,

Wavelength of wave is given as:

$λ = \frac{c}{v}$

$= \frac{3 \times 10^{8}}{50 \times 10^{2}} = 6.0 m$

(b) Assume the wave is travelling in a positive direction. The electric field vector will then point in the positive direction, and the magnetic field vector will also point in the positive direction. This is due to the fact that all three vectors are perpendicular to one another.

Equation of electric field vector is given as:

$\bar{E} = E_{0} \sin (k c - (i) t) j$

$= 120 \sin [1.05 x - 3.14 \times 10^{1} t] j$

And, magnetic field vector is given as:

$\bar{B} = B_{0} \sin (k x - (t) t) \hat{k}$

$\vec{B} = (4 \times 10^{- 1}) \sin [1.05 x - 3.14 \times 10^{9} t] \hat{k}$

Q. A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius $R = 6.0 cm$ has a capacitance $C = 100 pF$ . The capacitor is connected to a $230 V$ ac supply with a (angular) frequency of $300 rad s^{- 1}$

(a) What is the rms value of the conduction current?

(b) Is the conduction current equal to the displacement current?

Ans: Given that ,

Radius of each circular plate, $R = 6.0 cm = 0.06 m$

Capacitance of a parallel plate capacitor, $C = 100 pF = 100 \times 10^{- 12} F$

Supply voltage, $V = 230 V$

Angular frequency, $ω = 300 rad s^{- 1}$

(a) Rms value of conduction current, $I = \frac{V}{X_{ℓ}}$

Where,

$= \frac{1}{ω C}$

$∴ I = V \times ω C$

$= 230 \times 300 \times 100 \times 10^{- 12}$

$= 6.9 \times 10^{- 6} A$

$= 6.9 HA$

Hence, the rms value of conduction current is 6.9 $μ A$

(b) Yes, the displacement current equal to conduction current

$B = \frac{μ_{e} r^{r}}{2 π R^{2}} I_{e}$

Where,

$μ_{0} =$ Free space permeability $= 4 π \times 10^{- 7} N A^{- 1}$

$10 =$ Maximum value of current $= \sqrt{2} l$

$r =$ Distance between the plates from the axis $= 3.0 cm = 0.03 m$

$∴ B = \frac{4 π \times 10^{- 7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{- 5}}{2 π > 0 {(06)}^{2}}$

$= 1.63 \times 10^{- 11} T$ Hence, the magnetic field at that point is $1.63 \times 10^{- 11} T$ .

Q. Figure 8.6 shows a capacitor made of two circular plates each of radius $12 cm$ , and separated by $5.0 cm$ . The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to $0.15 A$

(a) Calculate the capacitance and the rate of charge of potential difference between the plates.

(b) Obtain the displacement current across the plates.

(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Ans: Given data,

Radius of each circular plate, $r = 12 cm = 0.12 m$

Distance between the plates, $d = 5 cm = 0.05 m$

Charging current, $I = 0.15 A$

Permittivity of free space, $ε_{0} = 8.85 \times 10^{- 12} C 2 N^{- 1} m^{- 2}$

(a) Capacitance between the two plates is given by the relation.

Ans: $C = \frac{ε_{0} A}{d^{'}} λ = \frac{c}{v} = \frac{3 \times 10^{8}}{30 \times 10^{6}} = 10 m$

Where,

$A =$ Area of each plate $= π r^{2}$

$C = \frac{ε_{0} π r^{2}}{d}$

$= \frac{8.85 \times 10 π^{12} \times 0 k 2 {()}^{2}}{0.05}$

$- 8.0032 \times 10^{- 12} F = 80.032 p F$

Charge on each plate, $q = C V$

Where,

$V$ = Potential difference across the plates

Differentiation on both sides with respect to time $(t)$ gives:

$\frac{d q}{d t} = C \frac{d V}{d t}$

But $\frac{d q}{a t} =$ current $(I)$

$∴ \frac{d V}{d t} = \frac{1}{C}$

$\Rightarrow \frac{0.15}{80.032 \times 10^{- 1.2}} = 1.87 \times 10^{9} V / s$

Therefore, the change in potential difference between the plates is $1.87 \times 10^{9} V / s$ .

(b) Obtain the displacement current across the plates.

Ans: The displacement current

(i d)

will be same as the conduction current or equal to the conduction current

(i c)

∴ (i d) = (i c) = 0.15 a m p

Ans: Yes, Kirchhoff’s first law of electrical circuits will be valid at each plate of the given capacitor, unless we consider the total sum of the both conduction and displacement currents.

Physics Class 12 Important Questions CBSE & State Board with Answers Unit – Optics

Q. State Huygens’s principle for constructing wave fronts.

Ans: According to Huygens’s principle:

Each source of light would spread waves in all directions.
Each point on the wave front would give rise to new disturbance which in-turn produces secondary wavelets which travel with the speed of light.
Only the forward envelope which encloses the tangent would give the new position of wave front.
Rays are always found to be perpendicular to the wave front.

Using Huygens’s principle, deduce the laws of reflection of light.

Ans: A plane wave front $A B$ incident at $A$ hence every point on $A B$ gives rise to new waves. Time taken by the ray to reach from $P$ to $R$

$t = \frac{P Q}{v} + \frac{Q R}{v}$ ……………….. (1)

In $Δ P A Q$ , $\sin i = \frac{P Q}{A Q}$

$\Rightarrow P Q = A Q \sin i$

In $Δ R C Q$ , $\sin r = \frac{Q R}{Q C}$

$\Rightarrow Q R = Q C \sin r$

Substituting in equation (1),

$\Rightarrow t = \frac{A Q \sin i}{v} + \frac{(A C - A Q) \sin r}{v}$

$\Rightarrow t = \frac{A Q \sin i}{v} + \frac{A C \sin r}{v} - \frac{A Q \sin r}{v}$

$\Rightarrow t = \frac{A Q (\sin i - \sin r)}{v} + \frac{A C \sin r}{v}$

Since all the secondary wavelets takes the same time to go from the incident wave front to the reflected wave front hence it must be independent of $Q$

i.e.,

sin i sin r = 0

$∴ \sin i = \sin r$

or $i = r$ (Law of Reflection of light)

Q. What changes in diffraction pattern of a single slit will you observe when the monochromatic source of light is replaced by a source of white light?

Ans: The changes would be:

1) The diffracted light consists of different colours.

2) It results in overlapping of different colours.

Q. Coloured spectrum is seen when we look through a muslin cloth. Why?

Ans: Muslin cloth is known to consist of very fine threads which acts as fine slits and when light passes through it, light would get diffracted giving rise to a coloured spectrum.

Q. What changes in diffraction pattern of a single slit will you observe when the monochromatic source of light is replaced by a source of white light?

Ans: The changes are:

Diffracted lights consist of different colours.
It would result in overlapping of different colours.

Q. A slit of width $^{'} a^{'}$ is illuminated by light of wavelength $6000 \overset{◯}{A}$ . For what value of $^{'} a^{'}$ will the:

First maximum fall at an angle of diffraction of $30^{\circ}$ ?

Ans: We are given:

$λ = 6000 \overset{◯}{A} = 6000 \times 10^{- 10} m$

$θ_{1} = 30^{\circ}, m = 1$

For first maximum

$\sin Q_{m} = \frac{(m + \frac{1}{2}) λ}{a}$

$\Rightarrow \sin Q_{1} = \frac{3 λ}{2 a}$

or $a = \frac{3 λ}{2 \sin θ_{1}} = \frac{3 \times 6 \times 10^{- 7}}{2 \times \sin 30^{\circ}}$

Q. First minimum fall at an angle of diffraction $30^{\circ}$ .

Ans: For first minimum

sin Q m = m λ a

Or $\sin Q_{1} = \frac{λ}{a}$

$\Rightarrow a = \frac{λ}{\sin θ_{1}}$

$\Rightarrow a = \frac{6 \times 10^{- 7}}{\sin 30^{\circ}}$

$∴ a = 1.2 \times 10^{- 5} m$

Q. Derive all expressions for the fringe width in Young’s double slit experiment.

Ans: Path difference between

$S_{1} P$ and $S_{2} P$

$Δ x = S_{2} P - S_{1} P$ …………………………… (A)

In $Δ S_{2} P$ ,

$\Rightarrow (S_{2} P) = {[{(S^{2} B)}^{2} + (P B^{2})]}^{\frac{1}{2}}$ …………………………… (1)

$S_{2} P = D {[1 + \frac{(y + \frac{d}{2})}{D^{2}}]}^{\frac{1}{2}}$

Using Binomial theorem expand equation $(1)$ and then neglect higher terms to get,

$S_{2} P = D + \frac{{(y + \frac{d}{2})}^{2}}{2 D}$

Similarly, $S_{1} P = D + \frac{{(y - \frac{d}{2})}^{2}}{2 D} \dots (2)$

Substituting equation (1) and (2) in equation $(A)$ ,

$\Rightarrow Δ x = \frac{{(y + \frac{d}{2})}^{2} - {(y - \frac{d}{2})}^{2}}{2 D}$

$\Rightarrow Δ x = \frac{y^{2} + {\frac{d}{4}}^{2} + y d - y^{2} - {\frac{d}{4}}^{2} + y d}{2 D}$

$Δ x = \frac{2 y d}{2 D}$

$Δ x = \frac{y d}{D}$

For bright fringes

Path difference $= x λ$

$x λ = \frac{y d}{D}$

\Rightarrow y = x λ D d

For $m = 1, y_{1} = \frac{λ D}{d}$

$n = 2, y_{2} = \frac{λ D}{d}$

For fringe width,

$β = y_{2} - y_{1}$

$∴ β = \frac{λ d}{d}$

Q. If the two slits in Young’s double slit experiment have width ratio $4 : 1$ deduce the ratio of intensity of maxima and minima in the interference pattern.

Ans: From the given ratio, $\frac{a_{1}^{2}}{a_{2}^{2}} = \frac{w_{1}}{w_{2}} = \frac{4}{1}$

a 1 a 2 = 2 1

a 1 = 2 a 2

Using

I max I min = ( a 1 + a 2 ) 2 ( a 1 - a 2 ) 2

\Rightarrow I max I min = ( 2 a 2 + a 2 ) 2 ( 2 a 2 - a 2 ) 2 = (3 a 2 a 2) 2

∴ I max I min = 9 1

Therefore, the ratio of intensity of maxima and minima in the interference pattern is $9 : 1$ .

Q. Monochromatic light of wavelength $589 n m$ is incident from air on a water surface. What are the wavelength, frequency and speed of :

Reflected, and (Refractive index of water is 1.33)

Ans: We are given,

Wavelength of incident monochromatic light, $λ = 589 n m = 589 \times 10^{- 9} m$

Speed of light in air, $c = 3 \times 10^{8} m / s$

Refractive index of water, $μ = 1.33$

The ray would reflect back in the same medium as that of the incident ray. Hence, the wavelength, speed, and frequency of the reflected ray will be the same as that of the incident ray.

Frequency of light is given by the relation,

$v = \frac{c}{λ}$

$\Rightarrow v = \frac{3 \times 10^{8}}{5000 \times 10^{- 10}} = 6 \times 10^{14} H z$

$∠ i + ∠ r = 90^{\circ}$

$\Rightarrow ∠ i + ∠ i = 90^{\circ}$

$\Rightarrow Z_{F} = \frac{{(4 \times 10^{- 3})}^{2}}{400 \times 10^{- 9}} = 40 m$

$\Rightarrow (λ^{'} - λ) = 15 \overset{◯}{A} = 15 \times 10^{- 10} m$

$\Rightarrow λ^{'} - λ = \frac{v}{c} λ$

$\Rightarrow v = \frac{c}{λ} \times (λ^{'} - λ)$

Also, $\frac{v}{c} = \frac{\sin i}{\sin r} = μ$

$\Rightarrow θ = 0.1^{\circ} = 0.1 \times \frac{λ}{180} = \frac{3.14}{1800} r a d$

And, $d = \frac{λ}{θ}$

$\Rightarrow d = \frac{600 \times 10^{- 9}}{\frac{3.14}{1800}} = 3.44 \times 10^{- 4} m$

$\Rightarrow λ = \frac{a^{2}}{Z_{F}}$

Now, we have, $n λ = x \frac{d}{D}$

$\Rightarrow d = \frac{n λ D}{d}$

$\Rightarrow θ = \frac{\frac{d}{d} λ}{d} = \frac{λ}{d}$

$\Rightarrow θ = \frac{3 \times 10^{8}}{589 \times 10^{- 9}}$

$∴ θ = 5.09 \times 10^{14} H z$

Hence, the speed, frequency, and wavelength of the reflected light are found to be $3 \times 10^{8} m$ , $5.09 \times 10^{14} H z$ , and $589 n m$ respectively.

Q. refracted light?

Ans: Frequency of light does not depend on the property of the medium in which it is travelling. So, the frequency of the refracted ray in water will be equal to the frequency of the incident or reflected light in air.

We have the refracted frequency to be $5.09 \times 10^{14} H z$

Speed of light in water is related to the refractive index of water as:

$V = \frac{c}{μ}$

$\Rightarrow V = \frac{3 \times 10^{8}}{1.33} = 2.26 \times 10^{8} m / s$

Wavelength of light in water is given by the relation,

$λ = \frac{V}{v}$

$\Rightarrow λ = \frac{2.26 \times 10^{8}}{5.09 \times 10^{14}}$

$\Rightarrow λ = 444.007 \times 10^{- 9} m$

$∴ λ = 444.01 n m$

Hence, the speed, frequency, and wavelength of refracted light are found to be $2.26 \times 10^{8} m / s$ , $444.01 n m$ and $5.09 \times 10^{14} H z$ respectively.

Q. In Young’s double-slit experiment using monochromatic light of wavelength $λ$ , the intensity of light at a point on the screen where path difference is $λ$ , is $K$ units. What is the intensity of light at a point where path difference is $λ / 3$ ?

Ans: Let $I_{1}$ and $I_{2}$ be the intensities of the two light waves. Their resultant intensities can be obtained as:

$I^{'} = I_{1} + I_{2} + 2 \sqrt{I_{1} I_{2}} \cos ϕ$

Where,

$ϕ =$ Phase difference between the two waves

For monochromatic light waves,

$I_{1} = I_{2}$

$\Rightarrow I^{'} = I_{1} + I_{1} + 2 \sqrt{I_{1} I_{1}} \cos ϕ$

$\Rightarrow I^{'} = 2 I_{1} + 2 I_{1} \cos ϕ$

Phase difference $= \frac{2 π}{λ} \times Path difference$

Since path difference $= λ$ ,

Phase difference, $ϕ = 2 π$

$∴ I = 2 I_{1} + 2 I_{1} = 4 I_{1}$

Given,

I' = K

$∴ I^{'} = \frac{K}{4}$ ………………………. (1)

When path difference $= \frac{λ}{3}$ ,

Phase difference, $ϕ = \frac{2 π}{3}$

Hence, resultant intensity, ${I_{R}}^{'} = I_{1} + I_{1} + 2 \sqrt{I_{1} I_{1}} \cos \frac{2 π}{3}$

$\Rightarrow {I_{R}}^{'} = 2 I_{1} + 2 I_{1} (- \frac{1}{2}) = I_{1}$

Using equation $(1)$ , we can write:

$I_{R} = I_{1} = \frac{K}{4}$

Therefore, the intensity of light at a point where the path difference is $\frac{λ}{3}$ is found to be $\frac{K}{4}$ units.

Physics Class 12 Important Questions State Board & CBSE with Answers Unit – Dual Nature of Radiation and Matter

Q. An electron and a proton are accelerated through the same potential. Which one of the two has
(i) greater value of de-Broglie wavelength associated with it and
(ii) less momentum? Justify your answer.

Answer:

Q. An electron and a photon each have a wavelength of 1.50 nm. Find
(i) their momenta,
(ii) the energy of the photon and
(iii) kinetic energy of the electron.
Answer:

Q. Draw a plot showing the variation of photoelectric current with collector plate potential for two different frequencies, v₁ > v₂, of incident radiation having the same intensity. In which case will the stopping potential be higher? Justify your answer.
Answer:
Stopping potential is directly proportional to the frequency of incident radiation. The stopping potential is more negative for higher frequencies of incident radiation. Therefore, stopping potential is higher in v₁.

Q. (a) Using de-Broglie’s hypothesis, explain with the help of a suitable diagram, Bohr’s second postulate of quantization of energy levels in a hydrogen atom.
(b) The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the state?
Answer:

Q. Write Einsten’s photoelectric equation. State clearly how this equation is obtained using the photon picture of electromagnetic radiation. Write the three salient features observed in photoelectric effect which can be explained using this equation. Answer:

This is Einstein’s photoelectric equation. Photoelectric emission is the result of interaction of two particles—one a photon of incident radiation and other an electron of photo sensitive metal. The free electrons are bound within the metal due to restraining forces on the surface. The minimum energy required to liberate an electron from the metal surface is called work function ϕ₀ of the metal. Each photon interacts with one electron. The energy hv of the incident photon is used up in two parts:
(a) a part of the energy of the photon is used in liberating the electron from the metal surface, which is equal to the work function ϕ₀ of the metal and
(b) the remaining energy of the photon is used in imparting K.E. of the ejected electron.
By the conservation of energy Energy of the inefficient photon = maximum K.E. of photoelectron + Work function

Three salient features are :
Three salient features observed in photoelectric effect on the basis of Einstein’s Photoelectric equation :

Q. Define the terms
(i) ‘cut-off voltage’ and
(ii) ‘threshold frequency’ in relation to the pheno-menon of photoelectric effect.
Using Einstein’s photoelectric equation show how the cut-off voltage and threshold frequency for a
given photosensitive material can be determined with the help of a suitable plot/graph.

Answer:
(i) Cut-off voltage : The value of the retarding potential at which the photo electric current becomes zero is called cut-off or stopping potential for the given frequency of the incident radiation.
(ii) Threshold frequency : The minimum value of the frequency of incident radiation below which the photoelectric emission stops altogether is called threshold frequency.
According to Eisntein’s photo electric equation,

Q. Draw a graph showing the variation of stopping potential with frequency of incident radiation for two photosensitive materials having work functions W₁ and W₂ (W₁ > W₂).
Write two important conclusions that can be drawn from the study of these plots.
Answer:

(i) Threshold frequency of material having work function W₁ is more than that of material of work function W₂.
(ii) The slopes of the straight line graphs, in both the cases, have the same value.
(iii) For the same frequency of incident radiation (> v₀₁), the maximum kinetic energy of the electrons, emitted from the material of work function W₁ is < that of electrons emitted from material of work function W₂.(any two)

Q. (a) Why photoelectric effect can not be explained on the basis of wave nature of light? Give reasons.
(b) Write the basic features of photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based.
Answer:
(a) (i) The maximum kinetic energy of the emitted electron should be directly proportional to the intensity of incident radiations but it is not observed experimentally. Also maximum kinetic energy of the emitted electrons should not depend upon incident frequency according to wave theory, but it is not so.
(ii) According to wave theory, threshold frequency should not exist. Light of all frequencies should emit electrons provided intensity of light is sufficient for electrons to eject.
(iii) According to wave theory, photoelectric effect should not be instantaneous. Energy of wave cannot be transferred to a particular electron but will be distributed to all the electrons present in the illuminated portion. Hence, there has to be a time lag between incidence of radiation and emission of electrons.

(b) Basic features of photon picture of electromagnetic radiation :
(i) Radiation behaves as if it is made of particles like photons. Each photon has energy E = hv and momentum p = h/λ.
(ii) Intensity of radiation can be understood in terms of number of photons falling per second on the surface. Photon energy depends only on frequency and is independent of intensity.
(iii) Photoelectric effect can be understood as the result of one to one collision between an electron and a photon.
(iv) When a photon of frequency
(v) is incident on a metal surface, a part of its energy is used in overcoming the work function and other part is used in imparting kinetic energy, so KE = h(v – v₀).

Q. Write Einstein’s photoelectric equation and point out any two characteristic properties of photons on which this equation is based. Briefly explain the three observed features which can be explained by this equation.
Answer:

Two characteristics properties of photons on which equation is based.
1. Photoelectric emission will take place only if frequency of incident radiation is greater than or equal to threshold frequency.
2. When a photon collides with an electron, it gives all its energy to electron.
Three features :

Q. (a) State three important properties of photons which describe the particle picture of electromagnetic radiation.
(b) Use Einstein’s photoelectric equation to define the terms
(i) stopping potential and
(ii) threshold frequency.
Answer:
(a)
Basic features of photon picture of electromagnetic radiation :
(i) Radiation behaves as if it is made of particles like photons. Each photon has energy E = hv and momentum p = h/λ.
(ii) Intensity of radiation can be understood in terms of number of photons falling per second on the surface. Photon energy depends only on frequency and is independent of intensity.
(iii) Photoelectric effect can be understood as the result of one to one collision between an electron and a photon.
(iv) When a photon of frequency
(v) is incident on a metal surface, a part of its energy is used in overcoming the work function and other part is used in imparting kinetic energy, so KE = h(v – v₀).

(b) (i) Stopping potential or cut-off potential. The minimum value of the negative potential ‘V₀‘, which should be applied to the anode in a photo cell so that the photo electric current becomes zero, is called stopping potential.
The maximum kinetic energy (K_max) of photoelectrons is given by,

(ii) Threshold frequency. The minimum frequency V₀, which the incident light must possess so as to eject photoelectrons from a metal surface, is called threshold frequency of the metal.

Q. An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

Answer:

For yellow light, wavelength X = 5.9 × 10^-7 m Since resolving power (R.P.) is inversely proportional to wavelength, therefore, R.P. of an electron microscope is about 10⁵ times more than optical microscope.

Q. Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation.
The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from λ₁ to λ₂. Derive the expressions for the threshold wavelength λ₀ and work function for the metal surface.
Answer:

(ii) Important features of photoelectric effect:
(a) Radiation behaves as if it is made of particles like photons. Each photon has energy E = hv and momentum p = h/λ.
(b) Intensity of radiation can be understood in terms of number of photons falling per second on the surface. Photon energy
depends only on frequency and is independent of intensity.
(c) Photoelectric effect can be understood as the result of the one to one collision between an electron and a photon.
(d) When a photon of frequency
(v) is incident on a metal surface, a part of its energy is used in overcoming the work function and other part is used in imparting kinetic energy, so KE = h(v – v₀)

Q. (a) Describe briefly three experimentally observed features in the phenomenon of photoelectric effect.
(b) Discuss briefly how wave theory of light cannot explain these features.

Answer:
(a) Experimental features and observations of photoelectric effect :
(i) For a given photosensitive material and frequency of incident radiation (above the threshold frequency), the photoelectric current is directly proportional to the intensity of incident light.
(ii) For a given photosensitive material and frequency of incident radiation, saturation current is found to be proportional to the intensity of incident radiation whereas the stopping potential is independent of its intensity.
(iii) For a given photosensitive material, there exists a certain minimum cut-off frequency of the incident radiation, called the threshold frequency, below which no emission of photoelectrons takes place, no matter how intense the incident light is. Above the threshold frequency, the stopping potential or equivalently the maximum kinetic energy of the emitted photoelectrons increases linearly with the frequency of the incident radiation, but is independent of its intensity.
(iv) The photoelectric emission is an instantaneous process without any apparent time lag (~10^-9 s or less), even when the incident radiation is made exceedingly dim.

(b) Wave theory cannot explain photoelectric effect:
(i) According to the wave picture of light, the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously. The greater the intensity of radiation, the greater are the amplitude of electric and magnetic fields. Consequently, the greater the intensity, the greater should be the energy absorbed by each electron. In this picture, the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with increase in intensity. Also, no matter what the ‘ frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons, so that they exceed the minimum energy needed to escape from the metal surface. A threshold frequency, therefore, should not exist. These expectations of the wave theory directly contradict observations (a) (i), (ii) and (iii) given above.

(ii) In the wave picture, the absorption of energy by electrons takes place continuously over the entire wavefront of the radiation. Since a large number of electrons absorb energy, the energy absorbed per electron per unit time turns out to be small. Explicit calculations estimate that it can take hours or more for a single electron to pick up sufficient energy to overcome the work function and come out of the metal. This conclusion is again in striking contrast to observation (iv) that the photoelectric emission is instantaneous.
In short, the wave picture is unable to explain the most basic features of photoelectric emission.

Q.(a) Write the important properties of photons which are used to establish Einstein’s photoelectric equation.
(b) Use this equation to explain the concept of
(i) threshold frequency and
(ii) stopping potential.
Answer:
(a) Important properties of Photons :
(i) In interaction of radiation with matter, radiation behaves as if it is made up of particles called photons.
(ii) Each photon has energy E (= hv) and momentum p (= hv/c), and speed c, the speed of light.
(iii) All photons of light of a particular frequency v, or wavelength λ, have the same energy E (= hv = hc/λ) and momentum p (= hv/c = h/λ), whatever the intensity of radiation may be. By increasing the intensity of light of given wavelength, there is only an increase in the number of photons per second crossing a given area, with each photon having the same energy. Thus, photon energy is independent of intensity of radiation.
(iv) Photons are electrically neutral and are not deflected by electric and magnetic fields.
(v) In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved. However, the number of photons may not be conserved in a collision. The photon may be absorbed or a new photon may be created.
(b) Einstein’s photoelectric equation is

This equation shows that the greater the work function ϕ₀, higher the threshold frequency v₀ needed to emit photoelectrons.
Thus, there exists a threshold frequency v₀ (=ϕ₀/h the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be or how long it falls on the surface.
(ii) Stopping potential. The minimum value of negative potential v₀, which should be applied to the anode in a photocell, so that the photoelectric current becomes zero, is called Stopping potential.

Q.Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only using Einstein’s equation.
Answer:
(a) (i) The maximum kinetic energy of the emitted electron should be directly proportional to the intensity of incident radiations but it is not observed experimentally. Also maximum kinetic energy of the emitted electrons should not depend upon incident frequency according to wave theory, but it is not so.
(ii) According to wave theory, threshold frequency should not exist. Light of all frequencies should emit electrons provided intensity of light is sufficient for electrons to eject.
(iii) According to wave theory, photoelectric effect should not be instantaneous. Energy of wave cannot be transferred to a particular electron but will be distributed to all the electrons present in the illuminated portion. Hence, there has to be a time lag between incidence of radiation and emission of electrons.