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

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Last updated date: 23rd Jun 2024
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Answer
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- Hint- Here, we will proceed by finding all the values included in the given data in SI units. Then, we will be using the formula for the kinetic energy acquired by a charge when it is revolving between the dees i.e., ${\text{KE}} = 2NqV$.

Complete step-by-step solution -

Given, Magnetic field, B = 2 T
Potential difference, V = 100 KV = 100$ \times $1000 V = ${10^5}$ V
Charge of any proton is simply charge corresponding to an electron i.e., q = e = $1.6 \times {10^{ - 19}}$ C
Kinetic energy, KE = 20 MeV = $20 \times {10^6}$ eV where e = $1.6 \times {10^{ - 19}}$ C (charge of a proton)
KE = $20 \times {10^6} \times 1.6 \times {10^{ - 19}}$ V
As we know that the kinetic energy acquired by a proton having charge q and potential difference of V after making N revolutions between the dees is given by
${\text{KE}} = 2NqV$
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}}} \\
   \Rightarrow N = 100 \\
 $
Therefore, a total of 100 revolutions are required to acquire a kinetic energy of 20 MeV.
Hence, option A is correct.

Note- In this particular problem, we are given with an extra data i.e., the magnitude of magnetic field which is equal to 2 teslas. This data is not at all required to solve this problem. In these types of problems, the units should be taken care of. We converted kilovolts into volts and MeV into volts in order to ensure the transparency of the units.