A proton and alpha particle both enter a region of uniform magnetic field, B, moving at right angles to the field B. If the radius of circular orbits for both the particles is equal and the kinetic energy acquired by proton is 1 MeV, the energy acquired by the alpha particle will be:

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Solution

Radius in magnetic field of circular orbit
$R = \frac{m v}{q B} = \frac{\sqrt{2 m E}}{q B}$

Total energy of a moving particle in a circular orbit,
$E = \frac{q^{2} B^{2} R^{2}}{2 m}$

For a proton enter in aregion of magnetic field,
$E = \frac{e^{2} \times B^{2} \times R^{2}}{2 \times m_{p}} . . . . . . . . . . . . (1)$

where $m_{p}$ is the mass of proton
Similarly for a $α$ particle in a uniform magnetic field

$E = \frac{(2 e)^{2} \times B^{2} \times R^{2}}{2 \times (4 m_{p})}$ $[m_{α} = 4 m_{p}] . . . . . . . . . . . . . . . . (2)$
Dividing eq(2) by (1) we get,

$\frac{E_{2}}{E_{1}} = \frac{(2 e)^{2} \times B^{2} \times R}{2 \times (4 m_{p})} \times \frac{2 \times m_{p}}{e^{2} \times B^{2} \times R^{2}}$
$\frac{E_{2}}{E_{1}} = 1$
$E_{2} = E_{1} = 1 M e v$

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