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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.

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When a charge is accelerated by a potential V, we have
`W = gV = 1/2 mv^(2) = (p^(2))/(2m)`
As work done on the charge becomes `KE` , so `P = sqrt(2mq^(V))`
`lambda = h/P rArr (lambda_(p))/(lambda_(alpha)) = sqrt((2xx 4m_(p) xx 2e xx v)/(2 xx m_(p) xx e xx v))rArr (lambda_(p))/(lambda_(alpha)) = (2sqrt(2))/(I)`.
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