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A proton is accelerating on a cyclotron ...

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

Text Solution

Verified by Experts

The correct Answer is:
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"`
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