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?

A proton and an alpha=particle are accelerated using the same potentia

1.	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?
Answer» Solution :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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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?

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