Question

A proton is accelerating in a cyclotron where the applied magnetic field is $2\;{\text{T}}$. If the potential gap effectively $100{\text{kV}}$ then how much revolutions the proton has to make between the "dees" to acquire a kinetic energy of $20{\text{MeV}}?$
(A) \[100\]
(B) \[150\]
(C) \[200\]
(D) \[300\]

Answer 1

Hint: In this question, we will proceed by finding all the values in SI units included in the given data. In order to directly get the required result, we will use a simple formula.
Formula Used: We will use the following formula for the kinetic energy acquired by a charge when it is revolving between the dees
${\text{KE}} = 2NqV$

Complete Step-by-Step Solution:
The following information is provided to us in the question
The applied magnetic field, $B = 2T$
The potential difference, $V = 100{\text{KV}} = 100 \times 1000\;{\text{V}} = {10^5}\;{\text{V}}$
The kinetic energy to be acquired is given as ${\text{KE}} = 20{\text{MeV}} = 20 \times {10^6}{\text{eV}}$
Where
$e = 1.6 \times {10^{ - 19}}{\text{C}}$(charge of a proton)
So, we get
${\text{KE}} = 20 \times {10^6} \times 1.6 \times {10^{ - 19}}\;{\text{V}}$
Now, we will use the formula listed above
${\text{KE}} = 2NqV$
By substituting the values of the kinetic energy, charge and the potential difference in the above equation, we get
$ \Rightarrow 20 \times {10^6} \times 1.6 \times {10^{ - 19}} = 2N\left( {1.6 \times {{10}^{ - 19}}} \right){10^5}$
$ \Rightarrow N = \dfrac{{20 \times {{10}^6} \times \left( {1.6 \times {{10}^{ - 19}}} \right)}}{{2\left( {1.6 \times {{10}^{ - 19}}} \right){{10}^5}}}$
Upon further solving, we get
$\therefore N = 100$

So, \[100\] revolutions are required to acquire a kinetic energy of \[20 MeV\]
Hence, the correct option is (A.)

Additional Information: A cyclotron is a type of accelerator for compact particles that produces radioactive isotopes that can be used for imaging procedures. In a magnetic field, stable, non-radioactive isotopes are placed into the cyclotron, which accelerates charged particles (protons) to high energy.

Note: In this specific problem, an additional data is given, i.e., the magnitude of the magnetic field equal to \[2 T\]. To solve this issue, this information is not at all necessary. The units should be taken care of in these types of challenges. In order to ensure the units' transparency, we converted kilovolts into volts and MeV into volts.