Question

An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii $r_e,r_p,r_{\alpha }$ respectively in a uniform magnetic field $B$. The relation between $r_e,r_p,r_{\alpha }$ is:

• A. $r_e<r_p<r_{\alpha }$
• B. $r_e<r_{\alpha }<r_p$
• C. $r_e>r_p=r_{\alpha }$
• D. $r_e<r_p=r_{\alpha }$

Answer 1

The correct option is D

$r_e<r_p=r_{\alpha }$

A charge particle in magnetic field

Force on particle due to magnetic field is,

$$\begin{gathered} F=qvB=\frac{m{v}^2}{r} \\ \Rightarrow r=\frac{mv}{qB} \end{gathered}$$

where $q$ is the charge of particle, $v$ is velocity, $B$ is the strength of magnetic field, $m$ is the mass and $r$ is the radius

Kinetic energy is,

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ \Rightarrow v=\sqrt{\frac{2K}{m}} \\ \Rightarrow r=\frac{m\sqrt{\frac{2K}{m}}}{qB}=\sqrt{\frac{2Km}{qB}} \end{gathered}$$

Since all three particles have same kinetic energy, we have $r\alpha \sqrt{\frac{m}{q}}$

$m_e$ is mass of electron, $m_{\alpha }$ mass of alpha particle and $m_p$is mass of proton

$m_{\alpha }=4m_p$and $m_e$is the smallest mass,

$m_e<m_p<m_{\alpha }$

Also charge of electron and proton is same but,$q_{\alpha }=4q_p$

$$\begin{gathered} \Rightarrow \frac{m_e}{q_e}<{\frac{m_p}{q_p}}_{}=\frac{m_{\alpha }}{q_{\alpha }} \\ \Rightarrow r_e<r_p=r_{\alpha } \end{gathered}$$

Hence, option D is correct.