A proton circulates in a cyclotron, beginning approximately at rest at the center. Whenever it passes through the gap between Dees, the electric potential difference between the Dees is 200 V.

(a)By how much does its kinetic energy increase with each passage through the gap?

(b)What is its kinetic energy as it completes 100passes through the gap? Let r₁₀₀be the radius of the proton’s circular path as it completes those 100passes and enters a dee, and let r₁₀₁be its next radius, as it enters a dee the next time.

(c)By what percentage does the radius increase when it changes from r₁₀₀to r₁₀₁? That is, what is Percentage increase =$\frac{{\mathbf{r}}_{\mathbf{101}}\mathbf{-}{\mathbf{r}}_{\mathbf{100}}}{{\mathbf{r}}_{\mathbf{100}}}\mathbf{100}\mathbf{\%}$?

• The potential difference between Dees = 200 eV
• r = r₁₀₀ when n = 100
• r = r₁₀₁ when n = 101

We are given the potential difference between Dees. According to the law of conservation of energy, when it passes through the gap, potential energy is converted into kinetic energy. From this, we can find the change in kinetic energy. We know the kinetic energy for completing one pass so that we can find the kinetic energy for 100 passes. Using the relation$\mathit{q}\mathit{v}\mathit{B}\mathbf{=}\frac{\mathbf{m}{\mathbf{v}}^{\mathbf{2}}}{\mathbf{r}}$, we will find the relation between r and n. Once we get that relation, we can find the percentage increase in the radius.

Formula:

role="math" localid="1662725924089" $\begin{array}{rcl}K& =& n\times K\\ qvB& =& \frac{m{v}^2}{r}\\ K{E}_1+P{E}_1& =& K{E}_2+P{E}_2\end{array}$

The total energy will always be conserved.

$\begin{array}{rcl}K{E}_1+P{E}_1& =& K{E}_2+P{E}_2\\ K{E}_2{-KE}_1& =& P{E}_2-P{E}_1\end{array}$

Potential energy will convert into kinetic energy, then:

$$\begin{gathered} KE=PE \\ KE=200eV \end{gathered}$$

Thus, 200 eV is the increase in kinetic energy.

As we know, the kinetic energy increase in one pass is 200 eV.

Kinetic Energy increase in 100 passes will be:

$\begin{array}{rcl}{K}_{100}& =& n\times K_{onepass}\\ K_{100}& =& 100\times 200\\ K_{100}& =& 20000eV\\ & =& 20.0keV\end{array}$

Thus, 20.0 keV is the kinetic energy as it completes 100 passes through the gap.

We know that:

$$\begin{gathered} K=\frac{1}{2}m{v}^2 \\ v=\sqrt{\frac{2K}{m}} \end{gathered}$$

We know that:

role="math" localid="1662726412882" $\begin{array}{rcl}qvB& =& \frac{m{v}^2}{r}\\ qB& =& \frac{mv}{r}\\ qB& =& \frac{m\left(\sqrt{{\displaystyle \frac{2K}{m}}}\right)}{r}\end{array}$

We can write this in terms of n as:

$\begin{array}{rcl}qB& =& \frac{m\left(\sqrt{{\displaystyle \frac{2nK}{m}}}\right)}{r}\\ r& =& \frac{m\left(\sqrt{{\displaystyle \frac{2nK}{m}}}\right)}{qB}\end{array}$

We can say that $\frac{m}{qB}\times \sqrt{\frac{2K}{m}}$is a constant term in the above equation.

So,

$R\propto \sqrt{n}$

From this, we can say that:

$$\begin{gathered} r_{100}\propto \sqrt{n_{100}} \\ {r}_{100}\propto \sqrt{n_{101}} \end{gathered}$$

$\begin{array}{rcl}Percentageincrease& =& \left|\frac{r_{101}-r_{100}}{r_{100}}\right|\times 100\\ & =& \left|\frac{\sqrt{n_{101}}-\sqrt{n_{100}}}{\sqrt{n_{100}}}\right|\times 100\\ & =& \left|\frac{\sqrt{101}-\sqrt{100}}{\sqrt{100}}\right|\times 100\\ & =& 0.499\%\end{array}$

Percentage increase in the radius when it changes from r₁₀₀ to r₁₀₁ will be 0.499 %.