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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.

Text Solution

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`lambda=h/"mv"=h/sqrt(2mE)=h/sqrt(2mqV)[because E=qV]`
For proton `m_p=m,q=e`
For `alpha`-particle `m_alpha`=4 m, q=2 e
`lambda_alpha/lambda_p=sqrt((m_pq_p)/(m_alphaq_alpha))`
`=1/(2sqrt2)`
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