Question

A proton, a deuteron and an $\alpha -particle$ with the same kinetic energy enter a region of uniform magnetic field moving at right angles to $B$. What is the ratio of the radii of their circular paths?

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

Answer 1

Answer: (B)

$1:\sqrt{2}:1$

$\frac{m{v}^2}{R}=qvB$
For proton, $R_p=\frac{mv}{Bq}=\frac{\sqrt{2M_{p}E}}{q_{p}B}$

Similarly for deuteron and $\alpha -particle$
$R_d=\frac{\sqrt{2M_{d}E}}{q_{d}B}$ and $R_{\alpha }=\frac{\sqrt{2M_{\alpha }E}}{q_{\alpha }B}$

According to the question
$\therefore R_p:R_d:R_{\alpha }$
or $\frac{\sqrt{M_p}}{q_p}:\frac{\sqrt{M_d}}{q_d}:\frac{\sqrt{M_{\alpha }}}{q_{\alpha }}$
$\therefore \frac{\sqrt{1}}{1}:\frac{\sqrt{2}}{1}:\frac{\sqrt{4}}{2}$
or $1:\sqrt{2}:1$.