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JEE Main 2022 (29 Jul, Shift 2) · Matter Waves: de Broglie Wavelength

Matter Waves: de Broglie Wavelength question from JEE Main 2022 (29 Jul, Shift 2)

Q82JEE Main 2022 · 29 JUL · SHIFT 2Easy

Matter Waves: de Broglie Wavelength

An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

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Answer: (b)

After acceleration, $K = qV$ and $p = \sqrt{2mK} = \sqrt{2mqV}$ , so $p \propto \sqrt{mq}$ .
$\dfrac{p_\alpha}{p_p} = \sqrt{\dfrac{4m_p \cdot 2e}{m_p \cdot e}} = \sqrt{8} = 2\sqrt{2}$
So the answer is (b) $2\sqrt{2} : 1$ .

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