Q85 / 90 · JEE Main 2021 (26 Aug Shift 2) · Physics

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:
$[m_{p} = 1.67 \times 10^{- 27} kg, e = 1.6 \times 10^{- 19} C$ , Speed of light $= 3 \times 10^{8} m s^{- 1}]$

Correct answer: 543

Solution

From the conservation of energy,

$2 n e V = \frac{1}{2} m v^{2}$

$\Rightarrow 2 n \times 12 \times 10^{3} \times 1.6 \times 10^{- 19} = \frac{1}{2} \times 1.67 \times 10^{- 27} \times {(5 \times 10^{7})}^{2}$

$\Rightarrow n = 543$

Hence, the correct answer is 543.