Question

A cyclotron's oscillation frequency is 10 MHz and the operating magnetic field is 0.66 T. If the radius of its dee is 60 cm, then the kinetic energy of the proton beam produced by the accelerator is ….A. 9 MeVB. 10 MeVC. 7 MeVD. 11 MeV

Hint: We will find out the mass of the proton from the given frequency and magnetic field. Then we will find an expression for the kinetic energy of particles coming out of the oscillator. Finally converting the energy into eV, we can have our answer.

Formula used:
$f=\dfrac{qB}{2\pi m}$
$E_k=\dfrac{q^2B^2R^2}{2m}$

Complete step by step solution:
The frequency of oscillation in a cyclotron is given by,
$f=\dfrac{qB}{2\pi m}$
Now, we know the frequency and the magnetic field B. From question, $B=0.66 T$ and $f=10 MHz$. Again, charge of a proton is equal and opposite to the charge of an electron and is given as $q=1.6 \times 10^{-19} C$ .
So, after putting all this values in the above formula, we obtain the mass of the proton as,
$m=\dfrac{qB}{2\pi f}=1.68\times 10^{-27}kg$
Now, the condition for uninterrupted oscillation in the cyclotron is that the magnetic force on the proton has to be equal to the centrifugal force of its rotation. So,
$qvB=\dfrac{mv^2}{r}$
Here, v is the velocity of the proton and r is the radius of its orbit. After a little modification the formula becomes, $mv=qBr$. It gives the momentum of the proton.
Now, when the proton is about to leave the dee, $r=R$. Here R is the radius of the dee and it’s about 60 cm or 0.6 m here. So, the final kinetic energy is given by,
$E_k=\dfrac{(mv)^2}{2m}=\dfrac{q^2B^2R^2}{2m}=1.19\times 10^{-12}J=7.47 MeV$
Hence, among the given options, option C is the closest to the actual value. Hence, option C is the correct answer.