A proton, an electron, a neutron and an alpha particle have the same energy. Their de Broglie's wavelengths are given below
(A) \( \lambda_{\mathrm{proton}} \)
(B) \( \lambda_{\mathrm{electron}} \)
(C) \( \lambda_{\mathrm{neutron}} \)
(D) \( \lambda_{\mathrm{alpha}} \)
Arrange these wavelengths in decreasing order.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B), (A), (C), (D)
The de Broglie wavelength of a particle of mass \( m \) and kinetic energy \( K \) is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \)
For the same energy, \( \lambda \propto 1/\sqrt{m} \): the lighter the particle, the longer its wavelength.
\( m_{e} \approx 9.1\times10^{-31}\,\mathrm{kg} \)
\( m_{p} \approx 1.67\times10^{-27}\,\mathrm{kg} \)
\( m_{n} \gt m_{p} \text{ (slightly)} \)
\( m_{\alpha} \approx 4m_{p} \)
The masses increase in the order electron, proton, neutron, alpha. So the wavelengths decrease in the order
\( \displaystyle \lambda_{\mathrm{electron}} \gt \lambda_{\mathrm{proton}} \gt \lambda_{\mathrm{neutron}} \gt \lambda_{\mathrm{alpha}} \)
that is (B), (A), (C), (D).
Note: This question is from the NCERT Exemplar. NTA sets numerical questions from the Exemplar too, so do not ignore it while practising numericals.
NCERT Exemplar:
Dual Nature of Radiation and Matter – …
