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 how many revolutions the proton has to make between the dees to acquire a kinetic energy of $20\text{MeV}$ ?

• A. $100$
• B. $250$
• C. $200$
• D. $300$

Answer 1

The correct option is A $100$

Kinetic energy of proton in joules is given by,

$K.E=20\text{MeV}=20\times {10}^6\text{eV}$

$\Rightarrow K.E=20\times {10}^6\times (1.6\times {10}^{-19})\text{J}$

The proton will get accelerated twice by the existing potential gap in dees during one complete revolution because both the dees of the cyclotron are identical.

The kinetic energy acquired by proton due to performing$N$ revolutions between the dees will be

$K.E=2(N\times qV)=2NqV$

$K.E=2N\times 1.6\times {10}^{-19}\times {10}^5$

$20\times {10}^6\times 1.6\times {10}^{-19}=2N\times 1.6\times {10}^{-19}\times {10}^5$

$\therefore N=100$

Hence, option $\left(a\right)$ is the correct answer.

Why this question ? Tip: During one complete revolution, the proton will cross the dees twice. Thus, it gets accelerated by existing potential difference two times in a complete revolution. Thus acquired $K.E=2N\left(qV\right)=2NqV$