Energy and momentum of a photon. h = 6.626e-34 J*s.
| Symbol | Quantity | SI Unit |
|---|---|---|
| E | energy | J |
| nu | frequency | Hz |
| lambda | wavelength | m |
| p | momentum | kg*m/s |
| h | Planck constant | J*s |
Try this first
Answer: B. The wave nature of light shows up in interference, diffraction and polarisation (NCERT Class 12 Physics Chapter 11, page 284).
A is wrong: A is wrong because the photoelectric and Compton effects involve energy and momentum transfer, where radiation behaves as a bunch of photons (NCERT Class 12 Physics Chapter 11, page 285).
C is wrong: C is wrong because it pairs the photoelectric effect, which needs the photon picture, with polarisation; only polarisation belongs to the wave side.
D is wrong: D is wrong because the Compton effect shows the particle nature; only diffraction belongs to the wave side.
A brighter beam of the same colour falls on a metal. Do the emitted electrons leave faster? No, and the reason is the whole topic. The wave picture says yes; the photon picture says only more electrons.
Wave picture first. Light is an electromagnetic wave with energy spread continuously over the wavefront, and interference, diffraction and polarisation show that wave nature (NCERT Class 12 Physics Chapter 11, page 284). If electrons absorbed this energy continuously, a more intense beam would give each electron more energy, a threshold frequency should not exist, and an electron could need hours to collect enough energy. Observation says the opposite on all three counts. Above the threshold frequency, emission is instantaneous, even for very dim radiation (page 280). The wave picture cannot explain the most basic features of photoelectric emission (page 281).
Photon picture. Radiation is built of quanta of energy hν. An electron absorbs one quantum. The intensity of light of a given frequency is the number of photons per second, so raising it raises the number of electrons per second, while the maximum kinetic energy is set by the energy of each photon (page 281). Each photon has energy E = hν and momentum p = hν/c, moves at speed c, and is electrically neutral. In a collision, energy and momentum are conserved, but the number of photons need not be (page 283).
So radiation is both: waves in interference and diffraction, particles in photoelectric and Compton effects. The nature of the experiment decides which description fits (pages 285 and 287). The same idea, run backwards, led de Broglie to give matter a wave character.
Watch-out: "intensity" and "photon energy" are different dials. Intensity changes the number of photons; frequency changes the energy of each.
Given
A photon and an electron each have an energy of 2.0 × 10¹ eV. Take c = 3.0 × 10⁸ m s⁻¹, the electron mass m = 9.0 × 10⁻³¹ kg and 1 eV = 1.6 × 10⁻¹⁹ J.
Required
The ratio of the linear momentum of the electron to that of the photon, p_e/p_ph.
Concept
A photon has momentum p = E/c (NCERT Class 12 Physics Chapter 11, page 283). A free electron's kinetic energy is E = p²/2m, so p = √(2mE). Both momenta enter the de Broglie relation as λ = h/p (page 285).
Formula
p_ph = E/c; p_e = √(2mE); p_e/p_ph = c√(2m/E).
Substitution
E = 2.0 × 10¹ × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹⁸ J. p_e/p_ph = 3.0 × 10⁸ × √(2 × 9.0 × 10⁻³¹ / 3.2 × 10⁻¹⁸).
Calculation
2 × 9.0 × 10⁻³¹ / 3.2 × 10⁻¹⁸ = 5.6 × 10⁻¹³, whose square root is 7.5 × 10⁻⁷. Then 3.0 × 10⁸ × 7.5 × 10⁻⁷ = 2.3 × 10². The 2 in 2mE is exact and does not limit the significant figures.
Final answer
p_e/p_ph = 2.3 × 10² (about 225), so at equal energy the electron has the larger momentum, and its de Broglie wavelength is the shorter one.
Common trap
Inverting the ratio, or using p = E/c for the electron as well. That formula belongs to the photon, whose energy is hν and which has no rest mass. The electron's momentum comes from E = p²/2m.
Similar NEET-style question
A photon and an electron each have an energy of 5.0 eV. What is p_e/p_ph? (Answer: E = 5.0 × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹⁹ J; p_e/p_ph = c√(2m/E) = 3.0 × 10⁸ × √(2 × 9.0 × 10⁻³¹ / 8.0 × 10⁻¹⁹) = 3.0 × 10⁸ × 1.5 × 10⁻⁶ = 4.5 × 10². Check: the energy is four times smaller than in the example and the ratio goes as 1/√E, so it is 2 × 225 = 450.)
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Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which group of phenomena shows the wave nature of light?
Answer: B. The wave nature of light shows up in interference, diffraction and polarisation (NCERT Class 12 Physics Chapter 11, page 284).
Why A is wrong: A is wrong because the photoelectric and Compton effects involve energy and momentum transfer, where radiation behaves as a bunch of photons (NCERT Class 12 Physics Chapter 11, page 285).
Why C is wrong: C is wrong because it pairs the photoelectric effect, which needs the photon picture, with polarisation; only polarisation belongs to the wave side.
Why D is wrong: D is wrong because the Compton effect shows the particle nature; only diffraction belongs to the wave side.
Which statement matches the chapter's conclusion about the wave and particle descriptions of radiation?
Answer: D. Radiation has a dual nature, wave and particle, and the nature of the experiment determines which description is best suited for understanding the result (NCERT Class 12 Physics Chapter 11, page 287).
Why A is wrong: A is wrong because the choice depends on the experiment, not on the lighting. Interference shows waves and the photoelectric effect shows photons in the same room.
Why B is wrong: B is wrong because a given experiment is explained by one description. Photoelectric emission needs photons, while interference is explained by waves (NCERT Class 12 Physics Chapter 11, pages 284 and 285).
Why C is wrong: C is wrong because the wave picture was not discarded. Interference, diffraction and polarisation remain explained by it (NCERT Class 12 Physics Chapter 11, page 284).
What is the energy of a photon of wavelength 5.0 × 10² nm? (Take h = 6.63 × 10⁻³⁴ J s and c = 3.0 × 10⁸ m s⁻¹.)
Answer: A. E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸)/(5.0 × 10⁻⁷) = 4.0 × 10⁻¹⁹ J, which is about 2.5 eV (NCERT Class 12 Physics Chapter 11, page 283, E = hν = hc/λ).
Why B is wrong: B is wrong because 2.5 is the energy in eV. The same number in joules would be 4.0 × 10⁻¹⁹ J.
Why C is wrong: C is wrong because 1.3 × 10⁻²⁷ is h/λ, the photon's momentum in kg m s⁻¹, not an energy in joules.
Why D is wrong: D is wrong because it divides by 5.0 × 10² m instead of 5.0 × 10⁻⁷ m: the wavelength was left in nanometres.
Which statement about photons is correct?
Answer: C. All photons of a given frequency have the same energy E = hν whatever the intensity; raising the intensity raises only the number of photons per second (NCERT Class 12 Physics Chapter 11, page 283).
Why A is wrong: A is wrong because photons are electrically neutral and are not deflected by electric and magnetic fields (NCERT Class 12 Physics Chapter 11, page 283).
Why B is wrong: B is wrong because in a photon-particle collision energy and momentum are conserved, but the number of photons may not be: a photon may be absorbed or a new one created (NCERT Class 12 Physics Chapter 11, page 283).
Why D is wrong: D is wrong because every photon moves with speed c, the speed of light, whatever its frequency (NCERT Class 12 Physics Chapter 11, page 283).
A photon and an electron (mass m) have the same energy E. The ratio (λ_photon/λ_electron) of their de Broglie wavelengths is: (c is the speed of light)
Answer: D. Each wavelength is h/p. For the photon p = E/c, so λ_photon = hc/E. For the electron E = p²/2m gives p = √(2mE), so λ_electron = h/√(2mE). The ratio is c√(2mE)/E = c√(2m/E) (NCERT Class 12 Physics Chapter 11, pages 283 and 285). This is an NTA question from 2025.
Why A is wrong: A is wrong because (1/c)√(E/2m) is the inverse of the answer; the photon, with p = E/c, has the longer wavelength at the same energy.
Why B is wrong: B is wrong because √(E/2m) has units of speed, not a pure number, and misses c. With λ = h/p for both, λ_photon/λ_electron = p_e/p_photon = √(2mE)/(E/c) = c√(2m/E).
Why C is wrong: C is wrong because c√(2mE) keeps E in the numerator. Dividing p_e = √(2mE) by p_photon = E/c gives c√(2mE)/E = c√(2m/E).
If c is the velocity of light in free space, the correct statements about photon among the following are: A. The energy of a photon is E = hν. B. The velocity of a photon is c. C. The momentum of a photon, p = hν/c. D. In a photon-electron collision, both total energy and total momentum are conserved. E. Photon possesses positive charge.
Answer: B. A, B, C and D are all stated in the photon picture: E = hν, speed c, p = hν/c, and conservation of total energy and momentum in a photon-electron collision. E is false because photons are electrically neutral (NCERT Class 12 Physics Chapter 11, page 283). This is an NTA question from 2024.
Why A is wrong: A is wrong because A and B only leaves out two true statements: each photon has momentum p = hν/c (C), and in a photon-electron collision the total energy and total momentum are conserved (D).
Why C is wrong: C is wrong because A, C and D only leaves out B, which is also true: each photon moves with speed c, the speed of light.
Why D is wrong: D is wrong because E is false: photons are electrically neutral and are not deflected by electric and magnetic fields. This option also leaves out C, p = hν/c.
Radiation of a fixed frequency above the threshold frequency falls on a metal. The intensity of the radiation is doubled. According to the photon picture, what happens to the number of photoelectrons emitted per second and to their maximum kinetic energy?
Answer: A. The intensity of light of a given frequency is set by the number of photons incident per second, so doubling it doubles the number of emitted electrons per second. The maximum kinetic energy is set by the energy of each photon, hν, which has not changed (NCERT Class 12 Physics Chapter 11, page 281).
Why B is wrong: B is wrong because the maximum kinetic energy is fixed by the energy hν of each photon, not by how many photons arrive. That is the wave-picture expectation, and observation contradicts it.
Why C is wrong: C is wrong because it swaps the two effects: intensity changes the number of photons per second, not the energy of each.
Why D is wrong: D is wrong because more photons per second means more electrons freed per second, so the number per second does increase.
Radiation of frequency 8.0 × 10¹⁴ Hz falls on a metal of work function 2.0 eV. What is the maximum kinetic energy of the photoelectrons? (Take h = 6.63 × 10⁻³⁴ J s and 1 eV = 1.6 × 10⁻¹⁹ J.)
Answer: C. The photon energy is hν = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.3 × 10⁻¹⁹ J = 3.3 eV. Each electron absorbs one quantum, so K_max = hν − φ₀ = 3.3 − 2.0 = 1.3 eV (NCERT Class 12 Physics Chapter 11, page 281).
Why A is wrong: A is wrong because 3.3 eV is the photon energy hν, before the work function needed to leave the metal is subtracted.
Why B is wrong: B is wrong because 2.0 eV is the work function itself, the minimum energy to escape, not the kinetic energy left over.
Why D is wrong: D is wrong because 5.3 eV adds the work function to the photon energy, where it must be subtracted: K_max = hν − φ₀.
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K_max = h ν - W, where K_max is maximum KE of emitted electron, hν is photon energy, W is work function of metal. Threshold ν₀ = W/h. Stopping potential V₀ satisfies eV₀ = K_max.
-- NCERT Class 12 Physics, Ch. 11, p. 279For 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
-- NCERT Class 12 Physics, Ch. 11, p. 280There exists a threshold frequency ν0 (= φ0/h) for the metal surface, below which no photoelectric emission is possible, no matter how intense the incident radiation may be. Above it, the photoelectric equation, Eq. (11.2), can be written as eV0 = hν - φ0 (11.4).
-- NCERT Class 12 Physics, Ch. 11, p. 282Energy E = hν = hc/λ. Momentum p = h/λ = E/c. Mass = 0 (rest mass). Travels at c. Number of photons = I·A/(hν).
-- NCERT Class 12 Physics, Ch. 11, p. 285Each photon has energy E (=hν) and momentum p (= h ν/c), and speed c, the speed of light Photons are electrically neutral and are not deflected by electric and magnetic fields In a photon-particle collision (such as photon-electron collision), the total energy and total momentum are conserved
-- NCERT Class 12 Physics, Ch. 11, p. 283Emission of electrons from a metal when light of suitable frequency falls on it. Discovered by Hertz, studied by Lenard. Below threshold frequency ν₀, no electrons emitted regardless of intensity.
-- NCERT Class 12 Physics, Ch. 11, p. 276Note that in all the above experiments, it is found that, if frequency of the incident radiation exceeds the threshold frequency, the photoelectric emission starts instantaneously without any apparent time lag, even if the incident radiation is very dim.
-- NCERT Class 12 Physics, Ch. 11, p. 280According 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 (i), (ii) and (iii) given at the end of sub-section 11.4.3. Further, we should note that in the wave picture, the absorption of energy by electron 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.
-- NCERT Class 12 Physics, Ch. 11, p. 281In 1905, Albert Einstein (1879-1955) proposed a radically new picture of electromagnetic radiation to explain photoelectric effect. In this picture, photoelectric emission does not take place by continuous absorption of energy from radiation. Radiation energy is built up of discrete units – the so called quanta of energy of radiation. Each quantum of radiant energy has energy hn, where h is Planck’s constant and n the frequency of light. In photoelectric effect, an electron absorbs a quantum of energy (hn ) of radiation. If this quantum of energy absorbed exceeds the minimum energy needed for the electron to escape from the metal surface (work function f0), the electron is emitted with maximum kinetic energy Kmax = hn – f0 (11.2) More tightly bound electrons will emerge with kinetic energies less than the maximum value. Note that the intensity of light of a given frequency is determined by the number of photons incident per second. Increasing the intensity will increase the number of emitted electrons per second. However, the maximum kinetic energy of the emitted photoelectrons is determined by the energy of each photon.
-- NCERT Class 12 Physics, Ch. 11, p. 281The dual (wave-particle) nature of light (electromagnetic radiation, in general) comes out clearly from what we have learnt in this and the preceding chapters. The wave nature of light shows up in the phenomena of interference, diffraction and polarisation.
-- NCERT Class 12 Physics, Ch. 11, p. 284photoelectric effect and Compton effect which involve energy and momentum transfer, radiation behaves as if it is made up of a bunch of particles – the photons. Whether a particle or wave description is best suited for understanding an experiment depends on the nature of the experiment. For example, in the familiar phenomenon of seeing an object by our eye, both descriptions are important. The gathering and focussing mechanism of light by the eye-lens is well described in the wave picture. But its absorption by the rods and cones (of the retina) requires the photon picture of light.
-- NCERT Class 12 Physics, Ch. 11, p. 285Radiation has dual nature: wave and particle. The nature of experiment determines whether a wave or particle description is best suited for understanding the experimental result.
-- NCERT Class 12 Physics, Ch. 11, p. 287No question in our NEET 2020–2025 set targets this topic directly.
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