A proton of mass m and charge +e is moving in | vedclass

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Question

(a) $r = \frac{{\sqrt {2mK} }}{{qB}}$$&rArr;$ $K \propto \frac{{{q^2}}}{m}$ $&rArr;$ $\frac{{{K_p}}}{{{K_\alpha }}} = {\left( {\frac{{{q_p}}}{{{q_\alp

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$1$

Vedclass · Answer

(a) $r = \frac{{\sqrt {2mK} }}{{qB}}$$&rArr;$ $K \propto \frac{{{q^2}}}{m}$ $&rArr;$ $\frac{{{K_p}}}{{{K_\alpha }}} = {\left( {\frac{{{q_p}}}{{{q_\alpha }}}} ight)^2} 	imes \frac{{{m_\alpha }}}{{{m_p}}}$
$&rArr;$$\frac{1}{{{K_\alpha }}} = {\left( {\frac{{{q_p}}}{{2{q_p}}}} ight)^2} 	imes \frac{{4{m_p}}}{{{m_p}}} = 1$ $==>$ ${K_\alpha } = 1\,MeV$.

A proton of mass $m$ and charge $+e$ is moving in a circular orbit in a magnetic field with energy $1\, MeV$. What should be the energy of $\alpha - $particle (mass = $4m$ and charge = $+ 2e),$ so that it can revolve in the path of same radius.......$MeV$

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