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What is cyclotron? Discuss the principle,construction,theory and working of a cyclotron.What is the

$\begin{array}{1 1} (a)\;E_{max}= \frac{q^2B^2}{2m}r^2_{max}\\ (b)\;E_{max}= \frac{qB}{2m}r^2_{max} \\ (c)\;E_{max}= \frac{qB}{2m}r_{max} \\ (d)\;E_{max}= \frac{q^2B^2}{r^2_{max}}\end{array} $

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A)
It is a device used to accelerate charged particles like protons, deutrons, $\alpha$-particles, etc, to very high energies. 

PRINCIPLE : 
A charged particle can be accelerated to very high energies by making it pass through a moderate electric field a number of times. This can be done with the help of a perpendicular magnetic field which throws the charged particle into a circular motion, the frequency of which does not depend on the speed of the particle and the radius of the circular orbit.

CONSTRUCTION :
As shown in figure, a cyclotron consists of the following main parts:

1. It consists of two small, hollow, metallic half-cylinders $D_1$ and $D_2$ called $dees$ as they are in the shape of $D$.
2. They are mounted inside a vacuum chamber between the poles of a powerful electromagnet. 
3. The dees are connected to the source of high frequency alternating voltage of few hundred kilovolts.
4. The beam of charged particles to be accelerated is injected into the dees near their centre, in a plane perpendicular to the magnetic field.
5. The charged particles are pulled out of the dees by a deflecting plate (which is negatively charged) through a window $W$.
6. The whole device is in high vacuum (pressure $\sim 10^{-6}$ mm of Hg) so that the air molecules may not collide with the charged particles. 

THEORY : 
Let a particle of charge $q$ and mass $m$ enter a region of magnetic field $\overrightarrow{B}$ with a velocity $\overrightarrow{v}$ normal to the field $\overrightarrow{B}$. The particle follows a circular path, the necessary centripetal force begin provided by the magnetic field. Therefore, 

Magnetic force on charge $q$ = Centripetal force on charge $q$
or
 $q v B \sin 90^o = \frac{mv^2}{r}$ or $r = \frac{mv}{qB}$

Period of revolution of the charged particle is given by 

$T = \frac{2 \pi r}{v} = \frac{2 \pi}{v} . \frac{mv}{qB} = \frac{2 \pi m}{qB}$

Hence frequency of revolution of the particle will be 

$f_c = \frac{1}{T} = \frac{qB}{2 \pi m} $

Clearly,this frequency is independent of both the velocity of the particle and the radius of the orbit and is called cyclotron frequency or magnetic resonance frequency.This is the key fact which is made use of in the operation of a cyclotron.

WORKING :
Suppose a positive ion,say a proton,enters the gap between the two dees and finds dee D1 to be negative.It gets accelerated towards dee $D_1$.As it enters the dee $D_1$,it does not experience any  electric field  due to shielding effect of the metallic dee. The perpendicular magnetic field throws it into a circular path.At the instant the proton comes out of dee $D_1$,,it finds dee $D_1$ positive and dee $D_2$. It moves faster through $D_2$ describing a larger semicircle than before. Thus if the frequency of the applied voltage is kept exactly the same as the frequency of revolution of the proton,then every time the proton reaches the gap between the two dees, the electric field is reversed  and proton receives a push and finally it acquires very high energy. This is called the cyclotron’s resonance condition. The proton follows a spiral path. The accelerated proton is ejected through a window by a deflecting voltage and hits the target.

MAXIMUM K.E OF THE ACCELERATED IONS : 
The ions will attain maximum velocity near the periphery of the dees. If $v_0$ is the maximum velocity acquired by the ions and $r_0$ is the radius of the dees, then

$\frac{mv_0^2}{r_0} = qv_0B$ or $v_0 = \frac{q B r_0}{m}$

The maximum kinetic energy of the ions will be 

$K_0 = \frac{1}{2} mv_0^2 = \frac{1}{2} m [\frac{q B r_0}{m}]^2 $
       or
$K_0 = \frac{q^2 B^2 r_0^2}{2m}$.

LIMITATIONS OF CYCLOTRON: 
1.According to the Einstein’s special theory of relativity,the mass of a particle increases with the increase in its velocity as 
                 $m = \frac{m_0}{\sqrt{1 - v^2 / c^2}}$
Where $m_o$ is the rest mass of the particle. At high velocities,the cyclotron frequency $(f_c=qB/2πm)$ will decrease due to increase in mass. This will throw the particles out of resonance with the oscillating field. That is,as the ions reach the gap between the dees, the polarity of the dees is not reversed at that instant. Consequently the ions are not accelerated further.
The above drawback is overcome either by increasing magnetic field as in a synchrotron or by decreasing the frequency of the alternating electric field as in a synchro-cyclotron.

2. Electrons cannot be accelerated in a cyclotron.A large increase in their energy increases their velocity to a very large extent.This throws the electrons out of step with the oscillating field.

3.Neutron,being electrically neutral,cannot be accelerated in a cyclotron.

USES OF CYCLOTRON: 
1.The high energy particles produced in a cylinder are used to bombard nuclei and study the resulting nuclear reactions and hence investigate nuclear structure.

2.The high energy particles are used to produce other high energy particles,such as Neutrons by collisions.These fast neutrons used in atomic reactions.

3.It is used to implant ions into solids and modify their properties or even synthesis new materials.

4.It is used to produce radioactive isotopes which are used in hospitals for diagnosis and treatment.





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