b
Given than $K_{p}=K_{a}$
We know that $r=\frac{m v}{q B}=\frac{(2 m K)^{1 / 2}}{q B}$
$\therefore r_{p}=\frac{\left(2 m_{p} K_{p}\right)^{1 / 2}}{q_{p} B}$ and $r_{\alpha}=\frac{\left(2 m_{\alpha} K_{\alpha}\right)^{1 / 2}}{q_{\alpha} B}$
Now $\frac{r_{p}}{r_{\alpha}}=\sqrt{\left(\frac{m_{p}}{m_{\alpha}}\right)} \times \frac{q_{\alpha}}{q_{p}}=\sqrt{\left(\frac{m_{p}}{4 m_{p}}\right)} \times \frac{2 e}{e}=1$