A proton of energy $8\ eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

Question

Dr. Abubacker · Accepted Answer

(c) $r = \frac{\sqrt{2mK}}{qB} \Rightarrow q \propto \sqrt{mK}\  \Rightarrow K \propto \frac{q^{2}}{m}$
$\Rightarrow \frac{K_{\alpha}}{K_{p}} = \left( \frac{q_{\alpha}}{q_{p}} ight)^{2} 	imes \frac{m_{p}}{m_{\alpha}} \Rightarrow \frac{K_{\alpha}}{8} = \left( \frac{2q_{p}}{q_{p}} ight)^{2} 	imes \frac{m_{p}}{4m_{p}} = 1$
$\Rightarrow K_{\alpha} = 8\ eV$

Q. A proton of energy $8\ eV$ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be

Explanation

Reviewed by Dr. Abubacker

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