A proton, a deuteron (nucleus of { }_1 H^2) a | vedclass

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(B) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.Since kinetic ener

Vedclass · Accepted Answer

$1: \sqrt{2}: 1$

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(B) The radius of a circular path of a charged particle in a magnetic field is given by $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$.Since kinetic energy $K$ and magnetic field $B$ are the same for all particles, we have $r \propto \frac{\sqrt{m}}{q}$.For a proton $(p)$: mass $m_p = m$, charge $q_p = q$. So, $r_p \propto \frac{\sqrt{m}}{q}$.For a deuteron $(d)$: mass $m_d = 2m$, charge $q_d = q$. So, $r_d \propto \frac{\sqrt{2m}}{q}$.For an $\alpha$-particle $(\alpha)$: mass $m_\alpha = 4m$, charge $q_\alpha = 2q$. So, $r_\alpha \propto \frac{\sqrt{4m}}{2q} = \frac{2\sqrt{m}}{2q} = \frac{\sqrt{m}}{q}$.Thus, the ratio $r_p : r_d : r_\alpha = \frac{\sqrt{m}}{q} : \frac{\sqrt{2m}}{q} : \frac{\sqrt{m}}{q} = 1 : \sqrt{2} : 1$.

$A$ proton, a deuteron (nucleus of ${ }_1 H^2$) and an $\alpha$-particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to the field. The ratio of the radii of their circular paths is:

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