Proton,deuteron,and alpha particle of same ki | vedclass

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(C) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv

Vedclass · Accepted Answer

$r_{\alpha} = r_p < r_d$

Vedclass · Answer

(C) The radius of a charged particle moving in a magnetic field is given by $r = \frac{mv}{qB}$.Since kinetic energy $K = \frac{1}{2}mv^2$,we have $mv = \sqrt{2mK}$.Substituting this into the radius formula: $r = \frac{\sqrt{2mK}}{qB}$.Since $K$ and $B$ are constant,$r \propto \frac{\sqrt{m}}{q}$.For a proton $(p)$: $m_p = m, q_p = e \Rightarrow r_p \propto \frac{\sqrt{m}}{e}$.For a deuteron $(d)$: $m_d = 2m, q_d = e \Rightarrow r_d \propto \frac{\sqrt{2m}}{e} = \sqrt{2} r_p$.For an alpha particle $(\alpha)$: $m_{\alpha} = 4m, q_{\alpha} = 2e \Rightarrow r_{\alpha} \propto \frac{\sqrt{4m}}{2e} = \frac{2\sqrt{m}}{2e} = r_p$.Comparing the values: $r_{\alpha} = r_p$ and $r_d = \sqrt{2} r_p \approx 1.414 r_p$.Thus,$r_d > r_{\alpha} = r_p$ is not explicitly listed,but checking the options,the relation $r_{\alpha} = r_p < r_d$ is correct.

Proton,deuteron,and alpha particle of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton,deuteron,and alpha particle are respectively $r_p, r_d$,and $r_{\alpha}$. Which one of the following relations is correct?

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