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 ($m_p=1.67\times {10}^{-27}kg$, $e=1.6\times {10}^{-19}C$)

• A. 13.36
• B. 12.52
• C. 14.89
• D. 14.49

Answer 1

Answer: (D)

14.49

Here, $v_c=11MHz=11\times {10}^{6}Hz$

$B=1T,R=55cm=55\times {10}^{-2}m,$

$q_e=1.6\times {10}^{-19}C;m_p=1.67\times {10}^{-27}kg,$

$\therefore K.E.=\frac{q^2{B}^2{R}^2}{2m}=\frac{(1.6\times {10}^{-19}{)}^2\times (1{)}^2\times (55\times {10}^{-2}{)}^2}{2\times 1.67\times {10}^{-27}}$

$=23.19\times {10}^{-13}J$

$K.E.=\frac{23.19\times {10}^{-13}}{1.6\times {10}^{-19}}=14.49\times {10}^{6}eV=14.49MeV$