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Question

The de-Broglie wavelength of a particle accelerated with a p.d of 150 V is 1010m. If it is accelerated with a p.d of 600 V, its wavelength will be

A
0.25A
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B
0.5A
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C
1.5A
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D
2A
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Solution

The correct option is B 0.5A
12mv2=qVmv=28Vm λ=hp=hmv=h2mqV λα1Vλ1λ2=V2V11010λ2= 600150=2 λ2=0.5A.

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