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A proton (mass $m$ and charge $+e$) and an $\alpha -$ particle (mass $4m$ and charge $+2e$) are projected with the same kinetic energy at right angles to the uniform magnetic field. Which one of the following statements will be true
The $\alpha -$ particle will be bent in a circular path with a small radius that for the proton
The radius of the path of the $\alpha -$ particle will be greater than that of the proton
The $\alpha -$ particle and the proton will be bent in a circular path with the same radius
The $\alpha -$ particle and the proton will go through the field in a straight line
Solution
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 $\frac{r_{p}}{r_{\alpha}}=\sqrt{\left(\frac{m_{p}}{m_{\alpha}}\right)} \times \frac{q_{\alpha}}{q_{p}}=\sqrt{\left(\frac{m_{p}}{4 m_{p}}\right)} \times \frac{2 e}{e}=1$
