A proton carrying $1 M e V$ kinetic energy is moving in a circular path of radius $R$ in uniform magnetic field. What should be the energy of an $α$ -particle to describe a circle of same radius in the same field?

1 M e V

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0.5 M e V

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4 M e V

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2 M e V

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Solution

The correct option is A $1 M e V$
For a particle of mass m and charge Q in a magnetic field, the radius of the path is given by,
$r = \frac{\sqrt{2} m (K E)}{q B}$
$q \propto \sqrt{m (K E)}$
$\frac{e}{2 e} = \sqrt{\frac{(m_{p}) (I M e V)}{(4 m_{p}) (K E)}}$
$\frac{1}{4} = \frac{1}{4 (K E)}$
$K E = 1 M e V$

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