An alpha particle, an electron and a gamma-ray photon have the same momentum. Which of them has the longest wavelength?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → None, all have the same wavelength
The de Broglie wavelength \( \lambda=h/p \) depends only on the magnitude of the momentum. The mass, speed and charge of the particle do not change it. The same relation holds for a photon, whose wavelength is also \( h/p \).
Option (1) The alpha particle — Incorrect. It is the heaviest, so it moves the slowest for the given momentum. But its wavelength \( h/p \) is the same as that of the others.
Option (2) The electron — Incorrect. It is light, so it moves much faster than the alpha particle. But with the same momentum, its wavelength \( h/p \) is also the same.
Option (3) The gamma-ray photon — Incorrect. A photon of momentum \( p \) has the wavelength \( \lambda=h/p \), equal to that of the two particles.
Option (4) None, all have the same wavelength — Correct. The alpha particle, the electron and the photon each have \( \lambda=h/p \), and with equal momentum all three have the same wavelength.
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”
\( \displaystyle \lambda=\frac{h}{p} \)\( \displaystyle =\frac{h}{mv} \qquad (11.5) \)
Wave Nature of Matter – “Equation (11.5) for a material particle is basically a hypothesis whose validity can be tested only by experiment. However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen,
\( p=h\nu/c \qquad (11.6) \)
Therefore,
\( \displaystyle \lambda=\frac{c}{\nu}=\frac{h}{p} \qquad (11.7) \)
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.”