A proton and an alpha particle have the same de-Broglie's wavelength. What is same for both of them?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Momentum
The de Broglie wavelength of a particle depends only on its momentum, \( \lambda = \frac{h}{p} \), where h is Planck's constant. Equal wavelengths therefore mean equal momentum.
\( \lambda_p = \lambda_\alpha \Rightarrow p_p = p_\alpha \)
Why the other options are incorrect
Option (1) Velocity — Incorrect. \( v = p/m \), and the α particle is about four times as massive as the proton, so its velocity is about one quarter of the proton's.
Option (2) Energy — Incorrect. The kinetic energy is \( p^{2}/2m \), so for equal momentum the lighter proton has about four times the energy of the α particle.
Option (3) Charge — Incorrect. The proton has charge \( +e \) and the α particle \( +2e \); the de Broglie wavelength does not depend on the charge.
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
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{mv} \qquad (11.5) \)”