A) 28 MHz
B) 280 MHz
C) 2.8 GHz
D) 28 GHz
Correct Answer: D
Solution :
Key Idea: In a cyclotron, the centripetal force is provided by the transverse magnetic field B, and the force on a particle travelling in a magnetic field (which causes it to curve) is equal to Bqv. In cyclotron, centripetal force = magnetic force i.e., \[\frac{m{{v}^{2}}}{r}=Bqv\] \[\therefore \] \[\frac{v}{r}=\frac{Bq}{m}\] where \[\frac{v}{r}=\omega ,\], therefore \[\omega =\frac{Bq}{m}\] and frequency \[f=\frac{\omega }{2\pi }=\frac{Bq}{2m\pi }\] Putting the numerical values from the question, we have \[B=1T,m=9.1\times {{10}^{-31}}kg,q=1.6\times {{10}^{-19}}C\] \[\therefore \] \[f=\frac{1.6\times {{10}^{-19}}\times 1}{2\times 3.14\times 9.1\times {{10}^{-31}}}=28\times {{10}^{9}}Hz\] \[\therefore \] \[f=28GHz\]
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