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Question

The de-Broglie wavelength of a proton (mass=1.6×1027kg) accelerated through a potential difference of 1 kV is :
  1. 600˚A
  2. 0.9×1012m
  3. 7˚A
  4. 0.9×1019nm

A
600˚A
B
0.9×1012m
C
7˚A
D
0.9×1019nm
Solution
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eV=KE=12mv2

KE=p22m

De-broglie wavelength,

λ=hp

λ=h2mE=h2mVe

Putting the values of constants, we get:

λ=6.6×10342×1.6×1027×1.6×1019×1000=0.9×1012m

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