Question

A cyclotron's oscillator frequency is $10 \, MHz$ and the radius of its dees is $60 \, cm$ , then the kinetic energy of the proton beam produced by the accelerator is [ $m_{proton}=1.67\times 10^{- 27} \, kg$ ]

• A. $9 \, MeV$
• B. $10 \, MeV$
• C. $7 \, MeV$
• D. $11 \, MeV$

Answer 1

Answer: (C)

$KE=\frac{q^{2} B^{2} R^{2}}{2 m}$
Given, $f=10MHz=10^{7} \, Hz$
$R=60 \, cm=0.6 \, m$
$f=\frac{q B}{2 \pi m}$
$\therefore \, \, KE=2\pi ^{2}mf^{2}R^{2}$
$=2 \times(3.14)^{2} \times 1.67 \times 10^{-27} \times 10^{14} \times 0.36 J$
$\approx7MeV$