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A proton carrying 1 MeV kinetic energy i...

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

1 MeV

B

0.5 MeV

C

4 MeV

D

2 MeV

Text Solution

Verified by Experts

The correct Answer is:
A
`r=(sqrt(2m(KE)))/(qB)`
`q prop sqrt(m(KE))`
`(e)/(2e)=sqrt(((m_(p))(1MeV))/((4m_(p))(KE)))`
`(1)/(4)=(1)/(4(KE))`
`KE=1MeV`
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