A proton, a deuteron (nucleus of ${ }_1 \mathrm{H}^2$ ) and an $\alpha$-particle with same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is :

Question

Aicharya · Accepted Answer

$r=\frac{m v}{q B}=\frac{\sqrt{2 m K}}{q B}$ $\Rightarrow \quad r \propto \frac{(m)^{1 / 2}}{q}$ $	herefore \quad r_p=\frac{\left(m_pight)^{1 / 2}}{q_p}, r_d=\frac{\left(m_dight)^{1 / 2}}{q_d}$ and $\quad r_\alpha=\frac{\left(m_\alphaight)^{1 / 2}}{q_\alpha}$ $	herefore \quad r_p: r_d: r_\alpha=\frac{\sqrt{m}}{q}: \frac{\sqrt{2 m}}{q}: \frac{\sqrt{4 m}}{2 q}$ $=1: \sqrt{2}: 1$

A proton, a deuteron (nucleus of ${ }_1 \mathrm{H}^2$ ) and an $\alpha$-particle with same kinetic energy…

Solution