A proton carrying 1, Me V kinetic energy is m | vedclass

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Question

Kinetic energy of a charged particle,

$K=\frac{1}{2} m v^{2} 	ext { or } v=\sqrt{\frac{2 K}{m}}$

Radius of the circular path of a charged parti

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Vedclass · Answer

Kinetic energy of a charged particle,

$K=\frac{1}{2} m v^{2} 	ext { or } v=\sqrt{\frac{2 K}{m}}$

Radius of the circular path of a charged particle in uniform magnetic field is given by

$R=\frac{m v}{B q}=\frac{m}{B q} \sqrt{\frac{2 K}{m}}=\frac{\sqrt{2 m K}}{B q}$

Mass of a proton, $m_{p}=m$

Mass of an $\alpha$ -particle, $m_{\alpha}=4 m$

Charge of a proton, $q_{p}=e$

Charge of an $\alpha$ -particle, $q_{\alpha}=2 e$

$	herefore \quad R_{p}=\frac{\sqrt{2 m_{p} K_{p}}}{B q_{p}}=\frac{\sqrt{2 m K_{p}}}{B e}$

and $R_{\alpha}=\frac{\sqrt{2 m_{\alpha} K_{\alpha}}}{B q_{\alpha}}=\frac{\sqrt{2(4 m) K_{\alpha}}}{B(2 e)}=\frac{\sqrt{2 m K_{\alpha}}}{B e}$

$	herefore \quad \frac{R_{p}}{R_{\alpha}}=\sqrt{\frac{K_{p}}{K_{\alpha}}}$

As $R_{p}=R_{\alpha}$ (given) $	herefore K_{\alpha}=K_{p}=1\, \mathrm{MeV}$

A proton carrying $1\, Me V$ 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 ?........$MeV$

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