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A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an $\alpha$-particle to describe a circle of same radius in the same field?
A
4 MeV
B
2 MeV
C
1 MeV
D
0.5 MeV
Detailed Solution
$R = \frac{mv}{qB} = \frac{\sqrt{2mE}}{qB}$
For equal radii: $\frac{\sqrt{2m_pE_p}}{q_pB} = \frac{\sqrt{2m_\alpha E_\alpha}}{q_\alpha B}$
$E_\alpha = \left(\frac{q_\alpha}{q_p}\right)^2\left(\frac{m_p}{m_\alpha}\right)E_p$, with $\frac{m_p}{m_\alpha} = \frac{1}{4}$ and $\frac{q_\alpha}{q_p} = 2$
$E_\alpha = (2)^2\times\frac{1}{4}\times E_p = E_p$
$E_\alpha = 1$ MeV
For equal radii: $\frac{\sqrt{2m_pE_p}}{q_pB} = \frac{\sqrt{2m_\alpha E_\alpha}}{q_\alpha B}$
$E_\alpha = \left(\frac{q_\alpha}{q_p}\right)^2\left(\frac{m_p}{m_\alpha}\right)E_p$, with $\frac{m_p}{m_\alpha} = \frac{1}{4}$ and $\frac{q_\alpha}{q_p} = 2$
$E_\alpha = (2)^2\times\frac{1}{4}\times E_p = E_p$
$E_\alpha = 1$ MeV