An alpha-particle of 1 text{ MeV} kinetic ene | vedclass

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(B) 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 $K$ is the

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$4$

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(B) 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 $K$ is the kinetic energy.From this,$K = \frac{r^2 q^2 B^2}{2m}$,which implies $K \propto \frac{q^2 r^2}{m}$.For an $\alpha$-particle,$q_{\alpha} = 2e$ and $m_{\alpha} = 4m_p$. For a proton,$q_p = e$ and $m_p = m_p$.Given $K_{\alpha} = 1 	ext{ MeV}$ and $r_{\alpha} = r$,we want to find $K_p$ for $r_p = 2r$.Using the ratio: $\frac{K_{\alpha}}{K_p} = \frac{q_{\alpha}^2 r_{\alpha}^2}{m_{\alpha}} 	imes \frac{m_p}{q_p^2 r_p^2}$.Substituting the values: $\frac{1}{K_p} = \frac{(2e)^2 (r)^2}{4m_p} 	imes \frac{m_p}{e^2 (2r)^2} = \frac{4e^2 r^2}{4m_p} 	imes \frac{m_p}{4e^2 r^2} = \frac{1}{4}$.Therefore,$K_p = 4 	ext{ MeV}$.

An $\alpha$-particle of $1 \text{ MeV}$ kinetic energy is moving on a circular path in a uniform magnetic field. The kinetic energy of a proton,to be moved in the same magnetic field on a circular path of double radius,is $..... \text{ MeV}$.

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