Question

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 of 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]

Hint:

A common mistake in cyclotron problems is to forget that the particle is accelerated twice per revolution. The energy gain per revolution is \(2qV_0\), not \(qV_0\). Calculations can sometimes be simpler if you convert the final kinetic energy to electron-volts (eV) first.

Answer 1

Correct Answer: 543

Step 1: Understanding the Question:
A proton in a cyclotron gains energy each time it crosses the gap between the dees. We are given the accelerating potential and the final desired speed. We need to find how many full revolutions it takes to reach that speed.
Step 2: Key Formula or Approach:
1. Calculate the final kinetic energy (K\(_f\)) of the proton using \( K = \frac{1}{2}mv^2 \).
2. In a cyclotron, the particle is accelerated twice in each revolution. The energy gained in one revolution is \( \Delta K_{rev} = 2 \times qV_0 \), where \(V_0\) is the accelerating potential.
3. The total number of revolutions (N) is the total kinetic energy gained divided by the energy gained per revolution: \( N = \frac{K_f}{\Delta K_{rev}} \).
Step 3: Detailed Explanation:
Given values:
Accelerating potential, V\(_0\) = 12 kV = 12 \( \times \) 10\(^3\) V
Proton mass, m\(_p\) = 1.67 \( \times \) 10\(^{-27}\) kg
Proton charge, e = 1.6 \( \times \) 10\(^{-19}\) C
Speed of light, c = 3 \( \times \) 10\(^8\) m/s
Final speed, v\(_f\) = c/6 = (3 \( \times \) 10\(^8\)) / 6 = 0.5 \( \times \) 10\(^8\) m/s
First, calculate the final kinetic energy (K\(_f\)) in Joules:
\[ K_f = \frac{1}{2} m_p v_f^2 = \frac{1}{2} (1.67 \times 10^{-27}) (0.5 \times 10^8)^2 \] \[ K_f = \frac{1}{2} (1.67 \times 10^{-27}) (0.25 \times 10^{16}) = 0.20875 \times 10^{-11} \text{ J} \] Next, calculate the energy gained per revolution in Joules:
\[ \Delta K_{rev} = 2eV_0 = 2 \times (1.6 \times 10^{-19} \text{ C}) \times (12 \times 10^3 \text{ V}) \] \[ \Delta K_{rev} = 38.4 \times 10^{-16} \text{ J} = 3.84 \times 10^{-15} \text{ J} \] Finally, calculate the number of revolutions (N):
\[ N = \frac{K_f}{\Delta K_{rev}} = \frac{0.20875 \times 10^{-11} \text{ J}}{3.84 \times 10^{-15} \text{ J}} \] \[ N = \frac{0.20875}{3.84} \times 10^4 \approx 0.05436 \times 10^4 = 543.6 \] The number of revolutions must be an integer. Since the question asks for the number of revolutions to achieve the speed, we take the integer part.
Step 4: Final Answer:
The number of revolutions made by the proton is 543.