Question

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let $r_p,r_e$ and $r_{He}$ be their respective radii, then

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

Answer 1

The correct option is B $r_e<r_p=r_{He}$

Given:
$E_{proton}=E_{electron}=E_{He}$

For circular orbital motion of particles in magnetic field we have,

$mvr=qvB\Rightarrow r=\frac{mv}{qB}$

As, $\frac{1}{2}m{v}^2=K.E.$

$\Rightarrow m^2{v}^2=2mK.E.$

$\Rightarrow mv=\sqrt{2mK.E.}]$

$\therefore r=\frac{\sqrt{2mK.E.}}{qB}$

For proton, electron and $\alpha -$ particle,

$m_{He}=4m_p$ and $m_p>>m_e$

Also $q_{He}=2q_p$ and $q_p=q_e$

$\therefore$ As KE of all the particles is same then,

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

$\therefore r_{He}=r_p>r_e$

Hence option $\left(B\right)$ is correct.