Question

(a) Draw a schematic sketch of a cyclotron. Explain clearly the role of crossed electric and magnetic field in accelerating the charge. Hence derive the expression for the kinetic energy acquired by the particles.
(b) An $\alpha$-particle and a photon are released from the centre of the cyclotron and made to accelerate.
(i) Can both be accelerated at the same cyclotron frequency ? Give reason to justify your answer.
(ii) When they are accelerated in turn, which of the two will have higher velocity at the exit slit of the dees ?

Answer 1

(a) Schematic sketch of cyclotron : The combination of crossed electric and magnetic field is used to increase the energy of the charged particle. Cyclotron uses the fact that the frequency of revolution of the charged particle in a magnetic field is independent of its energy. Inside the dees, particle is shielded from the electric field and magnetic field acts on the particle and makes it to go round in a circular path. Every time the particle moves from one dee to the other it comes under the influence of the electric field which ensures to increase the energy of the particles as the sign of the electric field changes alternately. The increase in energy increases the radius of the circular path so the accelerated particle moves in a spiral path.


Since, the magnetic force provides required centripetal force
$\therefore qvB=\frac{m{v}^2}{r}$
or $r=\frac{mv}{qB}$
Here, r is the radius of semi-circular path which a particle covers inside a dee.
Hence, the kinetic energy of ions,
$K.E.=\frac{1}{2}m{v}^2$
$K.E.=\frac{1}{2}m\frac{r^2{q}^2{B}^2}{m^2}[\because v=\frac{rqB}{m}]$
$K.E.=\frac{1}{2}\frac{r^2{q}^2{B}^2}{m^2}$
(b) (i) Let us consider :
m $\to$ mass of proton
q $\to$ charge of proton
$\therefore$ Mass of alpha particle = 4m
Charge of alpha particle = 2q
Cyclotron frequency is given by,
$v=\frac{Bq}{2\pi m}$
or $v\propto \frac{q}{m}$
for proton :
Frequency, $v_p\propto \frac{q}{m}$
For alpha particle,
Frequency, $v_{\alpha }\propto \frac{2q}{4m}$
or $v_{\alpha }\propto \frac{q}{2m}$
Thus, particles will not accelerate with same cyclotron frequency. The frequency of proton is twice than the frequency of alpha particle.
(ii) Velocity, $v=\frac{Bqr}{m}$
or $v\propto \frac{q}{m}$
For proton :
Velocity, $v_p\propto \frac{q}{m}$
For alpha particle :
Velocity, $v_{\alpha }\propto \frac{2q}{4m}$ or $v_{\alpha }\propto \frac{q}{2m}$
Thus, particles will not exit the dees with same velocity. The velocity of proton is twice the velocity of alpha particle.