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Question

An electron has the kinetic energy 2.8×1023J. De-Broglie wavelength will be nearly

(me=9.1×1031kg)


A

9.28 × 10-4 m

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B

9.28 × 10-7 m

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C

9.28 × 10-8 m

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D

9.28 × 10-10 m

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Solution

The correct option is C

9.28 × 10-8 m


From De-Broglie wave equation, we get

λ = hp λ = hmv

&K.E=12mv2 v = 2KEm

Now,λ = hλ

&v=2KEm

λ = h2mKE

λ =6.62×10242×9.1×1031×2.8×1023

λ=9.28 × 108m


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