The de-Brogile wavelength associated with an electron accelerated by a potential of 64 V is :

0.153 n m

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0.613 n m

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1.22 n m

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0.302 n m

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Solution

The correct option is D $0.153 n m$
The de-Brogile wavelength associated with voltage is given by
$λ = \frac{1.227}{\sqrt{v}} n m$
$= \frac{1.227}{\sqrt{64}} n m$
$= 0.153 n m$

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