Question

A cyclotron’s oscillating frequency is$10MHz$. The radius of dee is$60cm$. What is the kinetic energy of proton produced by the accelerator, if the magnetic field is$0.65Tesla$?

Answer 1

Step 1: Given data

Oscillating frequency of the cyclotron, $f=10MHz={10}^{7}Hz$

Radius of the dee, $R=60cm=0.6m$

Strength of the magnetic field, $B=0.65T$

Kinetic energy of the proton, $E_p=?$

Step 2: Assumptions

Mass of the proton, $M=1.67\times {10}^{-27}Kg$

Charge on the proton, $q=1.6\times {10}^{-19}C$

$1MeV=1.6\times {10}^{-13}J$

Tangential velocity of the proton$=v$

Angular velocity of the proton$=\omega$

Step 3: Formula used

$v=r\omega$

$v=r\times 2\pi f$ ……………………...(a)

$E_p=\frac{1}{2}\times m{v}^2$ …………………(b)

Step 4: Calculation of the kinetic energy of the proton

Substituting the given values in equation (a), we get

$v=0.6m\times 2\times 3.14\times {10}^{7}Hz$

$v=3.78\times {10}^{7}m/s$

Substituting the given values and the value of velocity obtained above in equation (b), we get

$$\begin{gathered} E_p=0.5\times 1.67\times {10}^{-27}kg\times {(3.78\times {10}^{7}m/s)}^2 \\ {E}_p=1.19\times {10}^{-12}J \end{gathered}$$

Converting the value of kinetic energy obtained in joules to million electron-volt, we get

$$\begin{gathered} E_p=\frac{1.19\times {10}^{-12}}{1.6\times {10}^{-13}}MeV \\ {E}_p=7.4MeV \end{gathered}$$

Hence, the kinetic energy of the proton is equal to $7.4MeV$.