An alfa particle and a proton have same velocity when they enter a uniform magnetic field.The period of revolution of alfa particle will be.........to that of proton?

Answer 1

Givens:

B is constant

v_p=v_α

q_p=1

q_α=2 (because alpha particle is 2 protons(+1) and 2 neutrons(0)

4m_p=m_α

Formulas

F_m=qvB

F_c=mv²/r

Because the magnetic force provides the centripetal force, you can set the above equal to each other and solve for radius.

qvB=mv²/r

r=mv/(qB)

Plug in for α and p

r_p=m_pv_p/(q_pB)

r_α=m_αv_α/(q_αB)

substitute in proton values in alpha equation

r_α=4m_pv_p/(2q_pB)

Simplify

r_α=2m_pv_p/(q_pB)

Therefore

r_α=2r_p

Since the radius of the alpha particle is twice the radius of the proton due to the difference in mass and charge, the Circumference of the alpha particle is twice the circumference of the proton (C=2πr)

Since the Period is how long it takes to complete one circle it equals C/v. Since they have the same tangential velocity, the Period of the alpha particle would be twice the period of the proton since the circumference is double.