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Question

A proton when accelerated through a potential difference of $$V$$, has a de Broglie wavelength $$\lambda$$ associated with it. If an alpha particle is to have the same de Broglie wavelength $$\lambda$$, it must be accelerated through a potential difference of:


A
V8
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B
V4
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C
4V
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D
8V
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Solution

The correct option is B $$\displaystyle\frac{V}{8}$$
Since the kinetic energy will be equal to the work done on particle,

$$K=qV$$

Also, $$p=\sqrt { 2mK } $$

Thus, $$\lambda =\dfrac { h }{ p } =\dfrac { h }{ \sqrt { 2mK }  } =\dfrac { h }{ \sqrt { 2mqV }  } $$

For both to have same $$\lambda$$,

$${ m }_{ p }{ q }_{ p }{ V }={ m }_{ a }{ q }_{ a }{ V }_{ a }$$

$${ V }_{ a }=\dfrac { 1 }{ (4)(2) } V=\dfrac { V }{ 8 } $$

Answer is option A.

Physics

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