If the maximum value of accelerating potential provided by a ratio frequency oscillator is

$\mathrm{V}=12 \mathrm{kV}$

Number of revolution $=\mathrm{n}$

$\mathrm{n}\left[2 \times \mathrm{q}_{\mathrm{P}} \times \mathrm{V}\right]=\frac{1}{2} \mathrm{~m}_{\mathrm{P}} \times \mathrm{v}_{\mathrm{P}}^{2}$

$\mathrm{n}\left[2 \times 1.6 \times 10^{-19} \times 12 \times 10^{3}\right.$

$=\frac{1}{2} \times 1.67 \times 10^{-27} \times\left[\frac{3 \times 10^{8}}{6}\right]^{2}$

$\mathrm{n}\left(38.4 \times 10^{-16}\right)=0.2087 \times 10^{-11}$

$\mathrm{n}=543.4$

Ans. 543