Question

A proton is accelerating on a cyclotron having oscillating frequency of 11 MHz in external magnetic field of 1 T. If the radius of its dees is 55 cm, then its kinetic energy (in MeV) is is `(m_(p)=1.67xx10^(-27)kg, e=1.6xx10^(-19)C)``
A. 13.36
B. 12.52
C. 14.89
D. 14.49

Answer 1

Correct Answer - D
Here, `upsilon_(c)="11 MHz "=11xx10^(6)Hz`
B = 1 T, R = 55 cm = `55xx10^(-2)m,`
`e=1.6xx10^(-19)C, m_(p)=1.67xx10^(-27)kg.`
`therefore" K.E. "=(q^(2)B^(2)R^(2))/(2m)=((1.6xx10^(-19))^(2)xx(1)^(2)xx(5xx10^(-2))^(2))/(2xx1.67xx10^(-27))`
`=23.19xx10^(-13)J`
`=(23.19xx10^(-13))/(1.6xx10^(-19))=14.49xx10^(6)" eV = 14.49 MeV"`