Question

An electron, a proton, a deuteron and an alpha particle, each having the same speed are in a magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles are respectively $R_e,R_p,R_d$ and $R_{\alpha }$. It follows that

• A. $R_e=2R_p$
• B. $R_p=R_d$
• C. $R_d=R_{\alpha }$
• D. $R_p=R_{\alpha }$

Answer 1

The correct option is C $R_d=R_{\alpha }$

When a charged particle is projected into a magnetic field with its velocity perpendicular to the direction of the field, it performs uniform circular motion with radius
$r=\frac{mv}{qB}$

i.e radius will depend on the value of the ratio $\frac{m}{q}$
Mass and charge values of the particles are given in the following table. (Assuming mass and charge of proton to be $m$ aand $e$ respectively)

	Proton	Deuteron	Alpha	Electron
mass	$m$	$2m$	$4m$	$\frac{m}{1836}$
charge	$e$	$e$	$2e$	$e$
$\frac{mass}{charge}$	$\frac{m}{e}$	$\frac{2m}{e}$	$\frac{2m}{e}$	$\frac{m}{1836e}$

From the ratio of $\frac{\text{mass}}{\text{charge}}$, we can see that
$R_d=R_{\alpha }$. Therefore option (c) is correct.