An early cyclotron at Cornell University was used from the 1930 s to the 1950 s to accelerate protons, which would then bombard various nuclei. The cyclotron used a large electromagnet with an iron yoke to produce a uniform magnetic field of \(1.3 \mathrm{T}\) over a region in the shape of a flat cylinder. Two hollow copper dees of inside radius \(16 \mathrm{cm}\) were located in a vacuum chamber in this region. (a) What is the frequency of oscillation necessary for the alternating voltage difference between the dees? (b) What is the kinetic energy of a proton by the time it reaches the outside of the dees? (c) What would be the equivalent voltage necessary to accelerate protons to this energy from rest in one step (say between parallel plates)? (d) If the potential difference between the dees has a magnitude of $10.0 \mathrm{kV}$ each time the protons cross the gap, what is the minimum number of revolutions each proton has to make in the cyclotron?

Step by step solution

01

Frequency of oscillation

We can determine the frequency of oscillation using the cyclotron equation: \begin{equation} f = \frac{qB}{2\pi m} \end{equation} where \(q\) is the charge of a proton, \(B\) is the magnetic field, and \(m\) is the mass of a proton. Given: Charge of proton, \(q = 1.6 \times 10^{-19} \;\text{C}\) Magnetic field, \(B = 1.3 \;\text{T}\) Mass of proton, \(m = 1.67 \times 10^{-27} \;\text{kg}\) Let's plug these values into the equation: \begin{equation} f = \frac{(1.6 \times 10^{-19} \;\text{C})(1.3 \;\text{T})}{2\pi(1.67 \times 10^{-27} \;\text{kg})} \end{equation} Calculating the frequency of oscillation: \begin{equation} f \approx 2.01 \times 10^{7} \;\text{Hz} \end{equation}

02

Kinetic energy of a proton

To find the kinetic energy of a proton, we can use the relationship between velocity, magnetic field, and radius of motion in a cyclotron: \begin{equation} v = \frac{qBr}{m} \end{equation} Now, we can use the following equation to find the Kinetic energy: \begin{equation} K = \frac{1}{2}mv^2 \end{equation} Calculating the final velocity: \begin{equation} v = \frac{(1.6 \times 10^{-19} \;\text{C})(1.3 \;\text{T})(0.16 \;\text{m})}{1.67 \times 10^{-27} \;\text{kg}} \approx 2.0 \times 10^{7} \;\text{m/s} \end{equation} Now, we can calculate the Kinetic energy: \begin{equation} K = \frac{1}{2}(1.67 \times 10^{-27} \;\text{kg})(2.0 \times 10^{7} \;\text{m/s})^2 \approx 3.35 \times 10^{-12} \;\text{J} \end{equation}

03

Equivalent voltage necessary to accelerate protons

To find the equivalent voltage, we can use the relationship between kinetic energy and voltage: \begin{equation} K = qV \end{equation} Solving for the voltage: \begin{equation} V = \frac{K}{q} = \frac{3.35 \times 10^{-12} \;\text{J}}{1.6 \times 10^{-19} \;\text{C}} \approx 2.09 \times 10^7 \;\text{V} \end{equation}

04

Minimum number of revolutions

To find the minimum number of revolutions, we can use the relationship between the voltage difference between the dees and the total voltage necessary to accelerate the protons: \begin{equation} n = \frac{V_{total}}{V_{dees}} \end{equation} Given: Potential difference between the dees, \(V_{dees} = 10.0 \times 10^3 \;\text{V}\) Calculating the minimum number of revolutions: \begin{equation} n = \frac{2.09 \times 10^7 \;\text{V}}{10.0 \times 10^3 \;\text{V}} \approx 2090 \end{equation} Therefore, the frequency of oscillation is \(2.01 \times 10^7 \;\text{Hz}\), the kinetic energy of a proton by the time it reaches the outside of the dees is \(3.35 \times 10^{-12} \;\text{J}\), the equivalent voltage necessary to accelerate protons to this energy from rest in one step is \(2.09 \times 10^7 \;\text{V}\), and the minimum number of revolutions each proton has to make in the cyclotron is 2090.

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