A proton and an alpha particle are accelerated through the same potential difference and enter a uniform magnetic field perpendicular to their velocity. The ratio of the radii of their circular paths is:

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$A proton and an alpha particle are accelerated through the same potential difference $V$ and enter a uniform magnetic field $B$ perpendicular to their velocity. The ratio of the radii of their circular paths $r_p / r_\alpha$ is:$

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Correct answer: C — $1:\sqrt{2}$

The radius of the path is given by

r = \frac{2 mV}{q B}

. Substituting the mass and charge values for a proton (

m, q

) and an alpha particle ($4m, 2q$), the ratio

r_{p} / r_{α} = \frac{m}{4 m} \times \frac{2 q}{q} = \frac{1}{2} \times 2 = 1

, wait, recalculating:

r_{p} / r_{α} = m /4 m \times (2 q / q) = (1/2) \times 2 = 1

. Actually, the ratio is

1 : 2

based on the charge-to-mass ratio.