Q129 / 155 · WBJEE 2022 · Physics

If the kinetic energies of an electron, an alpha particle and a proton having same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$ and $\varepsilon_3$ respectively, then

Correct answer
A
$\varepsilon_1>\varepsilon_3>\varepsilon_2$
B
$\varepsilon_1=\varepsilon_2=\varepsilon_2$
C
$\varepsilon_1 < \varepsilon_3 < \varepsilon_2$
D
$\varepsilon_1>\varepsilon_2>\varepsilon_3$

Solution

$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}, \sqrt{\mathrm{mk}}=\mathrm{const}$ $\mathrm{k} \propto \frac{1}{\mathrm{~m}}, \varepsilon_1>\varepsilon_3>\varepsilon_2$

Hence, the correct answer is option: 'A'.

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