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

Question:

If the maximum value of accelerating potential provided by a ratio frequency oscillator is $12 \mathrm{kV}$. The number of revolution made by a proton in a cyclotron to achieve one sixth of the speed of light is

$\left[\mathrm{m}_{\mathrm{p}}=1.67 \times 10^{-27} \mathrm{~kg}, \mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\right.$

Speed of light $\left.=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]$

Solution:

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

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

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