Question

A proton carrying 1 MeV kinetic energy is moving in a circular path of radius R in uniform magnetic field. What should be the energy of an α - particle to describe a circle of same radius in the same field?
1. 2 MeV
2. 1 MeV
3. 0.5 MeV
4. 4 MeV

Answer 1

Correct Answer - Option 2 : 1 MeV

CONCEPT:

• Kinetic energy(K.E): It is the energy that it possesses due to its motion.
  • It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

Formula:

K.E = 1/2 mv²

\(v = \sqrt {\frac{{2K.E}}{m}} \)

where m = mass of object , v = velocity of object(m/s)

Radius of charged particle in magnetic field :

\(R = \frac{{mv}}{{Bq}} = \frac{m}{{Bq}}\sqrt {\frac{{2K.E}}{m}} = \sqrt {\frac{{2mK.E}}{{Bq}}} \)

where, m = mass , q = charge B = magnetic flux density.

CALCULATION:

Given that, K.E of proton  = 1 MeV, Radius = R

From the above discussion 

\(R = \frac{{mv}}{{Bq}} = \frac{m}{{Bq}}\sqrt {\frac{{2K.E}}{m}} = \sqrt {\frac{{2mK.E}}{{Bq}}} \)

Mass of a proton, m_p = m

Mass of an α-particle. m_α = 4m

Charge of a proton, q_p = e

Charge of an α -particle, q_α = 2e

Substituting all the values in the above formula

\(R = \sqrt {\frac{{2{m_p}K.{E_p}}}{{B{q_p}}}} = \sqrt {\frac{{2{m_p}K.{E_p}}}{{Be}}} \)

 and, \(R = \sqrt {\frac{{2{m_α }K.{E_α }}}{{B{q_α }}}} = \sqrt {\frac{{2(4m)K.{E_α }}}{{B(2e)}}} = \frac{{\sqrt {2mK.{E_α }} }}{{Be}}\)

\(\therefore \frac{{{R_p}}}{{{R_α }}} = \sqrt {\frac{{K.{E_p}}}{{K.{E_α }}}} \)

As R_p = R_α = given , then K_α = K_p = 1 MeV

The correct option is 2.