If an electron and a photon have energy, then,
(A) Electrons and photons both have wavelengths
(B) Electron and photon both have momentum
(C) Both electron and photon do not have momenta
(D) The electron has energy and momentum and the photon has energy only
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (B) only
A photon of energy \( E \) carries momentum \( p = E/c \), and a moving electron has a de Broglie wavelength \( \lambda = h/p \).
(A) Electrons and photons both have wavelengths. — Correct. The photon has \( \lambda = hc/E \) and the electron has \( \lambda = h/p \).
(B) Electron and photon both have momentum. — Correct. The electron has \( p = mv \) and the photon has \( p = E/c = h/\lambda \).
(C) Both electron and photon do not have momenta. — Incorrect. Both carry momentum, as shown in (B).
(D) The electron has energy and momentum and the photon has energy only. — Incorrect. A photon has momentum too, even though its rest mass is zero.
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{mv} \qquad (11.5) \)
where m is the mass of the particle and v its speed. … However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen,
\( \displaystyle p = h\nu/c \qquad (11.6) \)
Therefore,
\( \displaystyle \frac{h}{p} = \frac{c}{\nu} = \lambda \qquad (11.7) \)
That is, the de Broglie wavelength of a photon given by Eq. (11.5) equals the wavelength of electromagnetic radiation of which the photon is a quantum of energy and momentum.”