A proton, an electron, and an alpha-particle | vedclass

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(B) The radius $R$ of a charged particle moving in a uniform magnetic field $B$ with velocity $v$ is given by $R = \frac{mv}{qB}$.Since $v$ and $B$ ar

Vedclass · Accepted Answer

$R_{\alpha} > R_p > R_e$

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(B) The radius $R$ of a charged particle moving in a uniform magnetic field $B$ with velocity $v$ is given by $R = \frac{mv}{qB}$.Since $v$ and $B$ are constant for all particles, $R \propto \frac{m}{q}$.For a proton $(p)$: $m_p = m$, $q_p = e$, so $R_p \propto \frac{m}{e}$.For an electron $(e)$: $m_e \approx \frac{m}{1836}$, $q_e = e$, so $R_e \propto \frac{m}{1836e}$.For an $\alpha$-particle: $m_{\alpha} \approx 4m$, $q_{\alpha} = 2e$, so $R_{\alpha} \propto \frac{4m}{2e} = 2\frac{m}{e}$.Comparing the ratios: $R_{\alpha} = 2(\frac{m}{e})$, $R_p = 1(\frac{m}{e})$, $R_e \approx 0.0005(\frac{m}{e})$.Thus, $R_{\alpha} > R_p > R_e$.

$A$ proton, an electron, and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$, and $R_{\alpha}$ are the radii of the circular paths of these particles, then:

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