Consider the following statements about the de Broglie wavelengths of a proton and an alpha particle. (Take the mass of the alpha particle as four times the mass of the proton.)
(A) When the two move with the same speed, the de Broglie wavelength of the proton is four times that of the alpha particle.
(B) When the two have the same kinetic energy, the de Broglie wavelength of the proton is twice that of the alpha particle.
(C) When the two have the same momentum, the alpha particle has the shorter de Broglie wavelength.
(D) When the two have the same kinetic energy, the proton has the larger momentum.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Only the momentum decides the de Broglie wavelength, \( \lambda=h/p \), or \( \lambda=h/mv \). For a particle of kinetic energy \( K \) the momentum is \( p=\sqrt{2mK} \), so
\( \displaystyle \lambda=\frac{h}{\sqrt{2mK}} \)
Here \( m_\alpha=4m_p \).
(A) When the two move with the same speed, the de Broglie wavelength of the proton is four times that of the alpha particle. — True. For the same \( v \), \( \lambda\propto1/m \):
\( \begin{aligned} \frac{\lambda_p}{\lambda_\alpha} &= \frac{m_\alpha}{m_p} \\ &= 4 \end{aligned} \)
(B) When the two have the same kinetic energy, the de Broglie wavelength of the proton is twice that of the alpha particle. — True. For the same \( K \), \( \lambda\propto1/\sqrt{m} \):
\( \begin{aligned} \frac{\lambda_p}{\lambda_\alpha} &= \sqrt{\frac{m_\alpha}{m_p}} \\ &= \sqrt{4} \\ &= 2 \end{aligned} \)
(C) When the two have the same momentum, the alpha particle has the shorter de Broglie wavelength. — False. With equal momenta, \( \lambda=h/p \) is the same for both. The larger mass of the alpha particle is balanced by its smaller speed.
(D) When the two have the same kinetic energy, the proton has the larger momentum. — False. At the same \( K \), \( p=\sqrt{2mK} \) is larger for the heavier particle. The alpha particle has twice the momentum of the proton. So its wavelength is the shorter one in (B).
NCERT (2026-27):
Wave Nature of Matter – “Clearly, from Eq. (11.5 ), λ is smaller for a heavier particle (large m) or more energetic particle (large v).”
\( \displaystyle \lambda=\frac{h}{p}=\frac{h}{mv} \qquad (11.5) \)
Summary – “9. The de Broglie wavelength (λ) associated with a moving particle is related to its momentum p as: λ = h/p.”