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.

Answer 1

(a) 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, the particle is acted upon only by magnetic field which makes it go round in a circular path. Every time the particle moves from one D to another it comes electric field acts upon it to increase it energy and its radius.

Radius of the trajectory $r=\frac{mv}{qB}$

Kinetic enegy of ions=$\frac{1}{2}m{v}^2$

$=\frac{1}{2}m\frac{r^2{q}^2{B}^2}{m}(\because \frac{rqB}{m}=v)$

(b)(i)Cyclotron frequency $\nu =\frac{Bq}{2\pi m}$

$\nu \propto \frac{q}{m}$

For proton,${\nu }_p\propto \frac{q}{m}$

For alpha particle, ${\nu }_{\alpha }\propto \frac{2q}{4m}\propto \frac{q}{2m}$

${\nu }_p=2{\nu }_{\alpha }$

Particles will not accelerate with same cylotron frequency.

(ii)Velocity, $v=\frac{Bqr}{m}$

For proton, $v_p\propto \frac{q}{m}$

For alpha particle $v_{\alpha }\propto \frac{2q}{4m}\propto \frac{q}{2m}$

$v_p=2v_{\alpha }$

Thus the particles wll not leave the dees with same velocity.