A proton, a deuteron and an alpha particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_{p}, r_{d}$ and $r_{α}$ denote respectively the radii of trajectories of these particles, then:

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$r_{p} = r_{d} = r_{α}$

$r_{α} = r_{p} < r_{d}$

$r_{α} = r_{d} > r_{p}$

$r_{α} > r_{d} > r_{p}$

answer is A.

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Detailed Solution

The radius of the circular path is given by

$r = \frac{m v}{q B} = \frac{\sqrt{2 m K}}{q B}, where K = \frac{1}{2} m v^{2}$

$Thus r \propto \frac{\sqrt{m}}{q} .$ Since K and B are the same for the three particles.

If m_pis the mass of a proton and $q_{p}$ its charge, then $m_{d} = 2 m_{p} and q_{d} = q_{p} and m_{α} = 4 m_{p}$ and $q_{α} = 2 q_{p}$

From these it follows that $r_{α} = r_{p} < r_{d}$ .

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