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A proton carrying $1 \mathrm{MeV}$ 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?
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Verified Answer
The correct answer is:
$1 \mathrm{MeV}$
For proton
$r=\frac{\sqrt{2 m(\mathrm{KE})}}{q B}$
So $\quad q \propto \sqrt{m(\mathrm{KE})}$
Hence
$$
\begin{aligned}
\frac{e}{2 e} & =\sqrt{\frac{\left(m_p\right)(1 \mathrm{MeV})}{\left(4 m_p\right)(\mathrm{KE})}} \\
\frac{1}{4} & =\frac{1 \mathrm{MeV}}{4 \mathrm{KE}} \\
\mathrm{KE} & =1 \mathrm{MeV}
\end{aligned}
$$
$r=\frac{\sqrt{2 m(\mathrm{KE})}}{q B}$
So $\quad q \propto \sqrt{m(\mathrm{KE})}$
Hence
$$
\begin{aligned}
\frac{e}{2 e} & =\sqrt{\frac{\left(m_p\right)(1 \mathrm{MeV})}{\left(4 m_p\right)(\mathrm{KE})}} \\
\frac{1}{4} & =\frac{1 \mathrm{MeV}}{4 \mathrm{KE}} \\
\mathrm{KE} & =1 \mathrm{MeV}
\end{aligned}
$$
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