Q 12-04-172JEE MainJEE Main 2021 (26 Aug, Shift 2)Hard
If the maximum value of accelerating potential provided by a radio frequency oscillator is $12$ kV. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is: [$m_p = 1.67\times10^{-27}$ kg, $e = 1.6\times10^{-19}$ C, Speed of light $= 3\times10^8$ m s$^{-1}$]
Numerical value type. Enter your answer.
Answer: 543
Final speed $v = \dfrac{c}{6} = 5\times10^7$ m s$^{-1}$.
$K = \frac{1}{2}m_pv^2 = \frac{1}{2}\times1.67\times10^{-27}\times2.5\times10^{15} = 2.09\times10^{-12}$ J.
Energy gained per revolution (two gap crossings) $= 2eV = 2\times1.6\times10^{-19}\times12000 = 3.84\times10^{-15}$ J.
$$n = \frac{2.09\times10^{-12}}{3.84\times10^{-15}} \approx 543$$
Solution by Sreeraj P, M.Sc Physics