Question

The oscillating frequency of a cyclotron is $ 10 \,MHz $ . If the radius of its Dees is $ 0.5 \,m $ , the kinetic energy of a proton, which is accelerated by the cyclotron is

• A. 10.2 MeV
• B. 2.55 MeV
• C. 20.4 MeV
• D. 5.1 MeV

Answer 1

The Correct Option is D

KE of charged possible in a cyclotron,
$ {{E}_{k}}=\frac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2m} $
But frequency $ f=\frac{qB}{2\pi m} $
$ \therefore $ $ {{E}_{k}}=\frac{{{(2\pi mf)}^{2}}{{r}^{2}}}{2m}=2{{\pi }^{2}}m{{f}^{2}}{{r}^{2}} $
Or $ {{E}_{k}}=2\times {{(3.14)}^{2}}\times 1.67\times {{10}^{-27}}\times {{(10\times {{10}^{6}})}^{2}} $
$ \times {{(0.5)}^{2}} $
$ =8.23\times {{10}^{-13}}J $
$ \therefore \,\,{{E}_{k}}\,=\frac{8.23\,\times {{10}^{-13}}}{1.6\,\times {{10}^{-19}}\,}\, $
$ =5.1\,\times {{10}^{6}}\,eV=5.1\,MeV $