The kinetic energy of an electron, $\alpha$-particle and a proton are given as $4 \mathrm{~K}, 2 \mathrm{~K}$ and $\mathrm{K}$ respectively. The de-Broglie wavelength associated with electron $(\lambda \mathrm{e}), \alpha$-particle $((\lambda \alpha)$ and the proton $(\lambda p)$ are as follows:

Question

$\lambda \alpha<\lambda p<\lambda e$

$\lambda \alpha>\lambda p>\lambda e$

$\lambda \alpha=\lambda p<\lambda e$

$\lambda \alpha=\lambda p>\lambda e$

Solution

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