The oscillating frequency of acyclotron is $10\, MHz$. If the radius of its dee is $0.5\, m$, the kinetic energy of a proton, which is accelerated by the cyclotron is......$MeV$
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$KE$ of charged particle in a cyclotron.
$E_{k}=\frac{q^{2} B^{2} r^{2}}{2 m}$
But frequency, $f=\frac{\mathrm{qB}}{2 \pi \mathrm{m}}$
$\therefore \mathrm{E}_{\mathrm{k}}=\frac{(2 \pi \mathrm{mf})^{2} \mathrm{r}^{2}}{2 \mathrm{m}}=2 \pi^{2} \mathrm{mf}^{2} \mathrm{r}^{2}$
or $\quad \mathrm{E}_{\mathrm{k}}=2 \times(3.14)^{2} \times 1.67 \times 10^{-27} \times\left(10 \times 10^{6}\right)^{2} \times(0.5)^{2}$
$=8.23 \times 10^{-13} \,\mathrm{J}$
$\therefore $ $\mathrm{E}_{\mathrm{k}}=\frac{8.23 \times 10^{-13}}{1.6 \times 10^{-19}}=5.1 \times 10^{6} \mathrm{eV}=5.1\, \mathrm{MeV}$
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