An electron and a photon each have the same de-Broglie wavelength of 1.0 nm. The ratio of their linear momenta is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1:1
The de Broglie relation \( \lambda = h/p \) applies to every particle, including the photon. So the momentum depends only on the wavelength:
\( \displaystyle p = \frac{h}{\lambda} \)
Equal wavelengths give equal momenta, whatever the mass of the particle, so the ratio is 1 : 1.
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
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{m v} \qquad (11.5) \)”
Wave Nature of Matter – “… 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,
\( \displaystyle \lambda = \frac{h}{p} = \frac{c}{\nu} \qquad (11.7) \)
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.”