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Question

A proton and an alpha particle are accelerated through the same potential difference. The ratio of wavelengths associated with proton and alpha particle respectively is :


A
1:22
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B
2:1
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C
22:1
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D
4:1
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Solution

The correct option is C $$2\sqrt{2}:1$$
We know, $$\dfrac{1}{2}mv^2=eV$$

$$mv = \sqrt{2meV}$$

$$wavelength, \  \lambda = \dfrac{h}{mv}$$

$$\lambda _p = \dfrac{h}{\sqrt{2m_{p}q_{p}V}}$$

$$\lambda _p = \dfrac{h}{\sqrt{2meV}}$$

$$\lambda _{\alpha} = \dfrac{h}{\sqrt{2m_{\alpha} q_{\alpha}V}}$$

$$= \dfrac{h}{\sqrt{2\times4m\times2e\times V}}$$

$$\dfrac{\lambda _p}{\lambda _{\alpha }}  = \dfrac{2\sqrt2}{1}$$

So, the answer is option (C).

Physics

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