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

A proton and an `alpha`=particle are accelerated using the same potential difference.How are the de-Broglie wavelengths `lambda_(p)` and `lambda_(a)` related to each other?

Text Solution

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According to formula ,`lambda=(h)/(p)=(1)/(sqrt(2mK))=(1)/(sqrt(2Vqm))`
Here V is same and so,
`therefore (lambda_(p))/(lambda_(a))=sqrt((q_(alpha)m_(alpha))/(q_(p)m_(p)))=sqrt(((2e)(4m_(p)))/((e)(m_(p))))=sqrt(8)=2sqrt(2)`
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