Question

A proton and an $\alpha$-particle, accelerated through the same potential difference, enter a region of uniform magnetic field normally. If the radius of the proton orbit is $10cm$, the radius of $\alpha$-orbit is

• A. $10cm$
• B. $10\sqrt{2}cm$
• C. $20cm$
• D. $5\sqrt{2}cm$

Answer 1

The correct option is B $10\sqrt{2}cm$

Radius of a particle's orbit $r=\frac{mv}{Bq}$

$\therefore$ $r=\sqrt{\frac{m^2{v}^2}{B^2{q}^2}}=\sqrt{\frac{2mqV}{B^2{q}^2}}$ where $\frac{1}{2}m{v}^2=qV$

$⟹$ $r\propto \sqrt{\frac{m}{q}}$

Given : $r_p=10$ cm $q_{\alpha }=2q_p$ $m_{\alpha }=4m_p$

$\therefore$ $\frac{r_{\alpha }}{r_p}=\sqrt{\frac{m_{\alpha }{q}_p}{m_p{q}_{\alpha }}}$

Or $\frac{r_{\alpha }}{10}=\sqrt{\frac{\left(4m_p\right)q_p}{m_p\left(2q_p\right)}}=\sqrt{2}$ $⟹r_{\alpha }=10\sqrt{2}$ cm