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An electron of mass m when accelerated t...

An electron of mass `m` when accelerated through a potential difference `V` has de - Broglie wavelength `lambda`. The de - Broglie wavelength associated with a proton of mass `M` accelerated through the same potential difference will be

A

`(lamda)/(3)sqrt((M)/(m))`

B

`(lamda)/(3).(M)/(m)`

C

`(lamda)/(3)sqrt((m)/(M))`

D

`(lamda)/(3).(m)/(M)`

Text Solution

Verified by Experts

The correct Answer is:
C
`lamda=(h)/(sqrt(2mqV))`
`:'` q is constant.
`:.lamdaprop(1)/(sqrt(mV))`
`(lamda_(p))/(lamda_(e))=sqrt((m_(e)V_(e))/(m_(p)V_(p)))=sqrt((mV)/(M9V))`
`(lamda_(p))/(lamda_(e))=sqrt((m)/(M))xx(1)/(3)`
`lamda_(p)=(lamda_(e))/(3)sqrt((m)/(M))=(lamda)/(3)sqrt((m)/(M))`.
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