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Question

Electrons are accelerated through a p.d. of 150V. Given m=9.1×1031 Js, the de Broglie wavelength associated with it is

A
1.5A0
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B
1.0A0
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C
3.0 A0
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D
0.5 A0
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Solution

The correct option is B 1.0A0
De-Brogtie wavelength
λ=hmv

And
12mv2=eV

mv=2meV

λ=hmv

λ=h2meV

λ=6.62×10342×9.1×1031×1.6×1019×150

λ=1.0 A0
So, the answer is option (B).

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