1.

A proton of energy `8 eV` is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will beA. `4 eV`B. `2 eV`C. `8 eV`D. `6 eV`

Answer» Correct Answer - C
`r=sqrt(2mK)/(qB)impliesqpropsqrtmKimpliesKprop(q^(2))/(m)`
`(K_(alpha))/(K_(p))=(q_(alpha)/(p_(p)))^(2)xx(m_(p))/(m_(alpha))implies(k_(alpha))/(8)=((2q_(p))/(q_(p)))^(2)xx(m_(p))/(4m_(p))=1`
` implies K_(alpha)=8eV`


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