Given:
ratio of energy is 2√2 : √3
Apply:
[tex]T=2\pi\sqrt[\frac{}{}]{\frac{mr}{qBv}}[/tex]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
