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A proton and helium nucleus are shot into a magnetic field at right angles to the field with same kinetic energy. Then the ratio of their radii is
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$1: 1$
The radius of the circular path, of a charges particle in magnetic field
$r=\frac{m v}{q B}=\sqrt{\frac{2 m E}{q B}}$
Here, kinetic energy for proton and helium is same and both are moving in the same magnetic field
$\therefore \quad r \propto \frac{\sqrt{m}}{q}$
So, $\frac{r_{P}}{r_{\mathrm{He}}}=\frac{\frac{\sqrt{m_{P}}}{q_{P}}}{\frac{\sqrt{m_{\mathrm{He}}}}{q_{\mathrm{He}}}}=\sqrt{\frac{m_{P}}{m_{\mathrm{He}}}} \times \frac{q_{\mathrm{He}}}{q_{P}}=\sqrt{\frac{m}{4 m}} \cdot \frac{2 q}{q}=\frac{1}{1}$
$r=\frac{m v}{q B}=\sqrt{\frac{2 m E}{q B}}$
Here, kinetic energy for proton and helium is same and both are moving in the same magnetic field
$\therefore \quad r \propto \frac{\sqrt{m}}{q}$
So, $\frac{r_{P}}{r_{\mathrm{He}}}=\frac{\frac{\sqrt{m_{P}}}{q_{P}}}{\frac{\sqrt{m_{\mathrm{He}}}}{q_{\mathrm{He}}}}=\sqrt{\frac{m_{P}}{m_{\mathrm{He}}}} \times \frac{q_{\mathrm{He}}}{q_{P}}=\sqrt{\frac{m}{4 m}} \cdot \frac{2 q}{q}=\frac{1}{1}$
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