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(A) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $m$ is the

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$\sqrt{2}$

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(A) The radius of a circular path for a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$,where $m$ is the mass,$K$ is the kinetic energy,$q$ is the charge,and $B$ is the magnetic field strength.Given that both particles have equal kinetic energy $K$ and enter the same magnetic field $B$,the ratio of their radii is $\frac{r_{d}}{r_{\alpha}} = \frac{\sqrt{2m_{d}K} / (q_{d}B)}{\sqrt{2m_{\alpha}K} / (q_{\alpha}B)} = \sqrt{\frac{m_{d}}{m_{\alpha}}} \cdot \frac{q_{\alpha}}{q_{d}}$.For a deuteron,mass $m_{d} = 2m_{p}$ and charge $q_{d} = e$. For an alpha particle,mass $m_{\alpha} = 4m_{p}$ and charge $q_{\alpha} = 2e$.Substituting these values: $\frac{r_{d}}{r_{\alpha}} = \sqrt{\frac{2m_{p}}{4m_{p}}} \cdot \frac{2e}{e} = \sqrt{\frac{1}{2}} \cdot 2 = \frac{1}{\sqrt{2}} \cdot 2 = \sqrt{2}$.

$A$ deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to

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