Discussion Forum

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 10 cm , the radius of $\alpha$-particle is 

Answer:

Explanation:

Radius of path   $r_{time}=\frac{1}{\beta}\sqrt{\frac{2mv}{q}}$

$\therefore$  $\frac{r_{\alpha}}{r_{p}}=\sqrt{\frac{m_{\alpha}}{m_{p}}}\sqrt{\frac{q_{p}}{q_{\alpha}}}$

or   $\frac{r_{\alpha}}{10}=\sqrt{\frac{4}{2}}\Rightarrow r_{\alpha}=10\sqrt{2}cm$