A) 28 MHz
B) 280 MHz
C) 2.8 GHz
D) 28 GHz
Correct Answer: D
Solution :
Cyclotron frequency \[v=\frac{Bq}{2\pi m}\] \[\Rightarrow v=\frac{1\times 1.6\times {{10}^{-19}}}{2\times 3.14\times 9.1\times {{10}^{-31}}}=2.79\times {{10}^{10}}{{H}_{z}}\] \[=27.9\times {{10}^{9}}HZ\cong \ 28\ GHZ\]
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