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Question

Write the expression for the de-Broglie wavelength associated with a charged particle having charge q and mass m, when it is accelerated by a potential V.

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Solution

Let v be the velocity gained by the given charged particle when it is accelerated through a potential difference of V volts.
Kinetic energy of the particle = 12mv2.
Kinetic energy of the particle = Work done on the particle by electric field.
12mv2=qV
or P22m=qV
p= 2mqV
de-Broglie wavelength (λ) associated with the particle,
λ=hp=h2mqV

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