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JEE Main 2023 (6 Apr, Shift 1) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2023 (6 Apr, Shift 1)

Q77JEE Main 2023 · 6 APR · SHIFT 1Medium

Matter Waves: de Broglie Wavelength

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

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Answer: (a)

$\lambda = \dfrac{h}{\sqrt{2mK}} \Rightarrow \lambda \propto \dfrac{1}{\sqrt{mK}}$ .
Electron: $m_e(4K)$ , tiny since $m_e \ll m_p$ . Proton: $m_pK$ . Alpha: $(4m_p)(2K) = 8m_pK$ .
Largest $mK$ gives smallest $\lambda$ : $\lambda_\alpha < \lambda_p < \lambda_e$ .
So the answer is (a).

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