What is the ratio of the kinetic energy if both a proton or an alpha particle moving with same velocity enters perpendicular to the B?

CONCEPT:

• A charged particle experiences a force when it moves in a magnetic field.

​⇒ F = qvBsinθ     -----(1)

Where, F = force due to magnetic field, q = magnitude of charge, v = velocity of charge, B = magnetic field, and θ = angle between v and B

• When the velocity of a charged particle is perpendicular to a magnetic field, it describes a circle and the radius of the circle is given by:

​\(⇒ r=\frac{mv}{qB}\)    

• Kinetic Energy: The energy possessed by a particle by the virtue of its motion is called kinetic energy. It is given by,

​\(⇒ KE=\frac{1}{2}m× v^{2}\)  

Where KE = kinetic energy, m = mass and v = velocity

CALCULATION:

Given:

• The radius of the circle is given by:

​\(⇒ r=\frac{mv}{qB}\)     -----(1)    

• The kinetic energy is given by,

​\(⇒ KE=\frac{1}{2}m× v^{2}\)     -----(2)

By equation 1 and equation 2,

\(\Rightarrow r=\sqrt{\frac{2km}{qB}}\)     -----(3)

By equation 3 the radius of the proton is given as,

\(\Rightarrow r_{p}={\frac{\sqrt{2km}}{qB}}\)     -----(4)

By equation 3 the radius of the deuteron is given as,

\(\Rightarrow r_{d}={\frac{\sqrt{2k\times2m}}{qB}}\)

\(\Rightarrow r_{d}=\sqrt{2}\times{\frac{\sqrt{2km}}{qB}}\)     -----(5)

By equation 3 the radius of the α-particle is given as,

\(\Rightarrow r_{\alpha}={\frac{\sqrt{2k\times4m}}{2qB}}\)

\(\Rightarrow r_{\alpha}={\frac{\sqrt{2km}}{qB}}\)     -----(6)

By equation 4, equation 5, and equation 6,

\(\Rightarrow r_{p}:r_{d}:r_{\alpha}\\\Rightarrow1:\sqrt{2}:1\)

Hence, option 4 is correct.