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 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$$^{-1}$$]

We have a cyclotron in which the proton is accelerated every time it passes through the gap between the two dees. The radio-frequency oscillator provides a maximum potential difference of 12 kV across the gap. Whenever the signs of the dees are opposite, the proton experiences an accelerating potential of this magnitude, gains kinetic energy, enters the circular dee, describes a semicircle, and then emerges at the gap again, where the potential has reversed so that it is once more accelerated. Hence, in one complete revolution the proton crosses the gap twice and is accelerated twice.

The energy gained in one single crossing is given by the elementary relation

$$\Delta E_{\text{cross}} = eV,$$

where $$e = 1.6 \times 10^{-19}\ \text{C}$$ is the charge on the proton and $$V = 12\ \text{kV} = 12 \times 10^{3}\ \text{V}.$$ Substituting the numbers,

$$\Delta E_{\text{cross}} = \bigl(1.6 \times 10^{-19}\bigr)\bigl(12 \times 10^{3}\bigr) = 1.6 \times 12 \times 10^{-19+3} = 19.2 \times 10^{-16}\ \text{J} = 1.92 \times 10^{-15}\ \text{J}.$$

Because the proton crosses the gap twice in one full revolution, the energy gained per revolution is

$$\Delta E_{\text{rev}} = 2 \,\Delta E_{\text{cross}} = 2 \times 1.92 \times 10^{-15}\ \text{J} = 3.84 \times 10^{-15}\ \text{J}.$$

The target speed of the proton is one sixth of the speed of light, i.e.

$$v = \frac{1}{6}\,c = \frac{1}{6}\,\bigl(3 \times 10^{8}\bigr)\ \text{m s}^{-1} = 5.0 \times 10^{7}\ \text{m s}^{-1}.$$

For this (non-relativistic) speed, the kinetic energy acquired is obtained from the classical formula

$$K = \frac{1}{2}mv^{2},$$

where $$m = 1.67 \times 10^{-27}\ \text{kg}$$ is the proton’s mass. Substituting,

$$K = \frac{1}{2}\,\bigl(1.67 \times 10^{-27}\bigr)\,\bigl(5.0 \times 10^{7}\bigr)^{2}.$$

First evaluate the square of the speed:

$$\bigl(5.0 \times 10^{7}\bigr)^{2} = 25 \times 10^{14} = 2.5 \times 10^{15}.$$

Now multiply:

$$K = \frac{1}{2} \times 1.67 \times 10^{-27} \times 2.5 \times 10^{15} = 0.835 \times 10^{-27} \times 2.5 \times 10^{15} = 2.0875 \times 10^{-12}\ \text{J}.$$

This is the total kinetic energy the proton must gain. Each revolution supplies $$\Delta E_{\text{rev}} = 3.84 \times 10^{-15}\ \text{J}.$$ Therefore the number of revolutions required is

$$N = \frac{K}{\Delta E_{\text{rev}}} = \frac{2.0875 \times 10^{-12}}{3.84 \times 10^{-15}} = \frac{2.0875}{3.84} \times 10^{3} \approx 0.5435 \times 10^{3} \approx 5.435 \times 10^{2}.$$

Keeping only whole revolutions, we need approximately 543 revolutions.

So, the answer is $$543$$.