Question

The oscillating frequency of a cyclotron is $10\text{MHz}$. If the radius of its dees is $0.5\text{m}$, then the kinetic energy of a proton accelerated by the cyclotron will be:

• A. $5.2\text{MeV}$
• B. $7.2\text{MeV}$
• C. $4.1\text{MeV}$
• D. $2.3\text{MeV}$

Answer 1

The correct option is A $5.2\text{MeV}$

Given:
$f=10\text{MHz}={10}^7\text{Hz};r=0.5\text{m}$

And for proton,
$m=1.67\times {10}^{-27}\text{kg};q=1.6\times {10}^{-19}\text{C}$

The kinetic energy of a particle accelerated by the cyclotron is given by

$K.E.=2{\pi }^{2}m{r}^2{f}^2$

Substituting the values for proton, we get,

$K.E.=2\times {\pi }^2\times 1.67\times {10}^{-27}\times (0.5{)}^2\times ({10}^7{)}^2$

$K.E.$ in electron volt $\left(\text{eV}\right)$ is given by:

$K.E.=\frac{20\times 1.67\times {10}^{-27}\times (25\times {10}^{-2}\times {10}^{14}}{1.6\times {10}^{-19}}\text{eV}$

$K.E.=5.21\times {10}^6\text{eV}$

$\therefore K.E.=5.2\text{MeV}$

Hence, option $\left(a\right)$ is the correct answer.