A) 10.2 MeV
B) 2.55 MeV
C) 20.4 MeV
D) 5.1 MeV
E) 1.5 MeV
Correct Answer: D
Solution :
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 \[{{E}_{k}}=2\times {{(3.14)}^{2}}\times 1.67\times {{10}^{-27}}\times {{(10\times {{10}^{6}})}^{2}}\] \[\times {{(0.5)}^{2}}\] \[=8.23\times {{10}^{-13}}J\] \[\therefore \,\,{{E}_{k}}\,=\frac{8.23\,\times {{10}^{-13}}}{1.6\,\times {{10}^{-19}}\,}\,=5.1\,\times {{10}^{6}}\,eV=5.1\,MeV\]
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