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Question

A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has
a) greater value of de-Broglie wavelength associated with it, and
b) less momentum?
Give reason to justify your answer.

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Solution

From law of conservation of energy, Kinetic energy equals the change in electrical energy in the particle.
12mv2=qV
p=mv=2mqV

De-broglie wavelength of a particle is given by: λ=hp
λ=h2mqV

(a) Ratio of de-broglie wavelength of proton to deuteron is given by:
λpλd=mdqdmpqp=2×11×1=2
Hence, proton has a higher de-broglie wavelength.

(b) p=hλ
Hence, deuteron has higher momentum.

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