The following particles have same kinetic energies. Arrange them in the descending order of their de-Broglie wavelengths.
A. β-particle
B. α-particle
C. Neutron
D. Proton
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A, D, C, B
The de Broglie wavelength of a particle with kinetic energy K is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \)
With the same K for all, \( \lambda \propto 1/\sqrt{m} \): the lighter the particle, the longer its wavelength.
\( m_{\beta} = 9.11\times10^{-31}\,\mathrm{kg} \text{ (an electron)} \)
\( m_{\mathrm{proton}} = 1.673\times10^{-27}\,\mathrm{kg} \)
\( m_{\mathrm{neutron}} = 1.675\times10^{-27}\,\mathrm{kg} \)
\( m_{\alpha} \approx 6.64\times10^{-27}\,\mathrm{kg} \)
Mass increases in the order β-particle, proton, neutron, α-particle, so the wavelength decreases in that order: A, D, C, B.