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Question

Electrons used in an electron microscope are accelerated by a voltage of 25 kV. If the voltage is increased to 100 kV then the de-Broglie wavelength associated with the electrons would :
  1. increase by 2 times
  2. decrease by 2 times
  3. decrease by 4 times
  4. increase by 4 times

A
decrease by 2 times
B
increase by 2 times
C
decrease by 4 times
D
increase by 4 times
Solution
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As the kinetic energy (K) of the electron is equal to the energy gained from accelerating through the electric potential (V), so

p22m=eV....(1)

The de-Broglie wavelength, λ=h/p

Now (1) becomes, h22mλ2=eV

so , λ1V

Thus, λ2λ1=V1V2=25/100=1/2

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