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A proton and a deuteron are accelerated ...

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 reasons to justify your answer.

Text Solution

Verified by Experts

(a) de-Broglie wavelength is given by
`lambda=(h)/(sqrt(2mqV))`
As mass of proton `lt` mass of deuteron and `q_(p)=q_(d)` and v is same
`implies lambda_(p) gt lambda_(d)` for same accelerating potential.
(b) Momentum `= (h)/(lambda)`
`:. lambda_(p) gt lambda_(d)`
`:.` Momentum of proton will be less, than that of deutron.
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