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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. 13.36B. 12.52C. 14.89D. 14.49 |
Answer» 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"` |
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