Question

An $\alpha$-particle and proton are accelerated from rest through same potential difference and both enter into a uniform perpendicular magnetic field. Find the ratio of their radii of curvature.

• A. $\sqrt{7}:1$
• B. $\sqrt{5}:1$
• C. $\sqrt{2}:1$
• D. $\sqrt{3}:1$

Answer 1

Answer: (C)

$\sqrt{2}:1$

We know,

$R=\frac{mv}{qB}$

$R=\frac{\sqrt{2mqV}}{qB}=\frac{\sqrt{2mV}}{\sqrt{q}B}$

Here,

B,V is constant for both cases,

$R\propto \frac{\sqrt{m}}{\sqrt{q}}$

Let $m_p=m,q_p=q$

Then,$m_{\alpha }=4m,q_{\alpha }=2q$

$R_p:R_{\alpha }=\frac{\sqrt{m}}{\sqrt{q}}:\frac{\sqrt{4m}}{\sqrt{2q}}=1:\sqrt{2}$

$R_{\alpha }:R_p=\sqrt{2}:1$

Option $\text{C}$ is the correct answer.