Question

A proton and an $\alpha$-particle enter a uniform magnetic field at right angles to the field with the same speed. The ratio of the periods of $\alpha$-particle and proton is-

• A. $2:1$
• B. $1:1$
• C. $1:4$
• D. $1:2$

Answer 1

The correct option is A $2:1$

Proton ${\to }_{}1H^1$ and $\alpha$-particle $\to {}_{2}H{e}^4$

$\Rightarrow q_p=e;q_{He}=2e;m_p=m;m_{He}=4m$

$\because \overrightarrow{v}\perp \overrightarrow{B}$ Hence, they will undergo a uniform circular motion.

$F_m=F_{centripetal}$

$qvB=\frac{m{v}^2}{r}[\because v=r\omega ]$

$Bqr\omega =mr{\omega }^2\Rightarrow Bq=m\omega$

$Bq=m\left(\frac{2\pi }{T}\right)$

$\therefore T=\frac{2\pi m}{Bq}\Rightarrow T\propto \frac{m}{q}$

$\frac{T_{\alpha }}{T_p}=\frac{m_{\alpha }}{m_p}\times \frac{q_p}{q_{\alpha }}=\frac{4}{1}\times \frac{1}{2}=\frac{2}{1}$

$\therefore T_{\alpha }:T_p=2:1$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.