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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 uniform magnetic field. What should be the energy of an `alpha-` particle to describe a circle of the same radius in the same field?

A

2 Me V

B

1 Me V

C

0.5 Me V

D

4 Me V

Text Solution

Verified by Experts

The correct Answer is:
B
`k_(p) = 1` MeV = `10^(6) xx 1.6 xx 10^(-19) J`
`m_(p) = m , m_(a) = 4m , q_(p) = q , q_(a) = 2q`
Since , `r = (sqrt(2mk))/(qB) implies k = (r^(2) q^(2)B^(2))/(2m) implies k prop (q^(2))/(m)`
`therefore (k_(p))/(k_(alpha)) = ((q_(p))/(q_(alpha)))^(2) xx ((m_(alpha))/(m_(p)))`
`implies (k_(p))/(k_(alpha)) = (1)/(4) xx 4 implies k_(a) = k_(p) implies k_(alpha) = 1 MeV`
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