Question

A proton is accelerating in a cyclotron where the applied magnetic field is $2\text{T}$ . If the potential gap is effectively $100\text{kV}$ , then find the number of revolutions made by the proton between the “dees” to acquire kinetic energy of $20\text{MeV}$ .
A) 100
B) 150
C) 200
D) 300

Answer 1

Hint: Both the “dees” of the cyclotron are identical, so one-half revolution will be completed in one of the “dees” while the other half will be completed in the other one. So the kinetic energy for one complete revolution will be twice the product of the charge of the proton and the potential gap between the “dees”. Then the number of revolutions to be done by the proton to obtain the given kinetic energy can be easily obtained.

Formula used:
-The kinetic energy of a charge in a cyclotron after $N$ number of revolutions is given by, $K=2NqV$ where $q$ is the charge and $V$ is the potential difference between the “dees” of the cyclotron.

Complete step by step solution:
Step 1: List the known parameters of the given problem.

The magnetic field applied is given to be $B=2\text{T}$ .
The potential difference maintained between the two “dees” is given to be $V=100\text{kV}=100\times {10}^3={10}^5\text{V}$ .
The charge of the proton will be the charge of the electron but positive i.e., $q=e=1\cdot 6\times {10}^{-19}\text{C}$
The kinetic energy acquired by the proton after $N$ number of revolutions is given to be $K_N=20\text{MeV}=20\times {10}^6\times \text{1}\cdot \text{6}\times \text{1}{\text{0}}^{-19}\text{V}$ .

Step 2: Express the kinetic energy acquired by the proton after one revolution.
The kinetic energy of the proton revolving in the given cyclotron after one revolution $\left(N=1\right)$ can be expressed as $K_1=2NeV=2eV$ ------- (1)
Substituting for $V=100\text{kV}$ and $e=1\cdot 6\times {10}^{-19}\text{C}$ in equation (1) we get the kinetic energy of the proton after one complete revolution as $K_1=2\times \left(1\cdot 6\times {10}^{-19}\right)\times {10}^5\text{V}$ .

Step 3: Using equation (1) obtain the number of revolutions made by the proton to acquire the given kinetic energy.
Equation (1) gives the kinetic energy of the proton after $N$ revolutions as $K_N=2NeV$ and that after one revolution as $K_1=2eV$
Then the number of revolution can be expressed as $N={\displaystyle \frac{K_N}{K_1}}$ ------- (2)
Substituting for $K_N=20\times {10}^6\times \text{1}\cdot \text{6}\times \text{1}{\text{0}}^{-19}\text{V}$ and $K_1=2\times \left(1\cdot 6\times {10}^{-19}\right)\times {10}^5\text{V}$ in equation (2) we get, $N={\displaystyle \frac{20\times {10}^6\times \left(\text{1}\cdot \text{6}\times \text{1}{\text{0}}^{-19}\right)\text{V}}{2\times \text{1}\cdot \text{6}\times \text{1}{\text{0}}^{-19}\times {10}^5\text{V}}}=100$
$\therefore$ The required number of revolutions is 100.

So, the correct option is A.

Note: The given value of the applied magnetic field is unnecessary for our calculations. While substituting values of physical quantities in an equation make sure that the transparency of units is maintained. If this is not the case, then the necessary conversion of units must be done. Here we converted the potential gap expressed in kilovolts into volts and the given kinetic energy of the proton expressed in mega electron-volts into volts before substituting their values in equations (1) and (2) respectively.