1.

A proton carrying `1 MeV` kinetic energy is moving in a circular path of radius `R` in unifrom magentic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?A. `2MeV`B. `1MeV`C. `0*5MeV`D. `4MeV`

Answer» Here, `K.E._("proton")=1MeV`, Also `K.E.=1/2mv^2`
So, `1/2mv^2=K.E`.,
So `mv^2=2K.E.implies mv=sqrt(2m(K.E.))`
Then `(mv^2)/(R)=Bqv R=(mv)/(Bq)=(sqrt(2m(K.E.)))/(Bq)`
So `K.E._P=(q^2pB^2R^2)/(2m_p)=(e^2B^2R^2)/(2m)`
For `alphararr q_alpha=2e, m_alpha=4m` proton
`K.E._alpha=(q_alpha^2B^2R^2)/(2m_alpha)=((2e)^2R^2B^2)/(2(4m))=(e^2R^2B^2)/(2m)`
`(K.E._alpha)/(K.E._("proton"))=1` or `K.E._alpha=1xxK.E._("proton")`
`=1xx1MeV=1MeV`


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