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A proton and an alpha-particle are accel...

A proton and an `alpha`-particle are accelerated through same potential difference. Find the ratio of their de-Brogile wavelength.

A

`1:1`

B

`1:2`

C

`2:1`

D

`2sqrt2 : 1`

Text Solution

Verified by Experts

The correct Answer is:
D
`qV = 1/2 mv^2 " or " mv = sqrt(2q Vm)` ,
So `lamda = (h)/(mv) = (h)/(sqrt(2 qVm) ) i.e. lamda prop (1)/(sqrt(qm)) ` ,
so ` lamda_p/lamda_alpha = sqrt( (q_alpha m_alpha)/(q_p m_p) ) = sqrt(2 xx 4) = 2sqrt2`
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