Question

Two protons enter a region of the transverse magnetic field. What will be the ratio of the time period of revolution if the ratio of energy is 2√2 : √3 ?

Answer 1

Given:

ratio of energy is 2√2 : √3

Apply:

$T=2\pi\sqrt[\frac{}{}]{\frac{mr}{qBv}}$

Where:

q = charge of proton

v= speed of proton

r= radius of circular path

T= time period of revolution

Kinetic energy (K)

K= 1/2mv^2

From both equations:

Tα1/k

K1:K2 = 2√2 : √3

T1:T2 = √3:2√2

Answer: √3:2√2