Question

If a proton is accelerated beyond the energy of $50\text{MeV}$, the cyclotron device begins to fail because :

• A. Frequency of revolution of proton increases steadily
• B. Assumption of frequency of revolution being independent of particle's speed does not hold, and at very high speed $(\sim 0.3c)$, proton gets out of step with cyclotron
• C. The corresponding size of magnet for a conventional cyclotron would be impossibly expensive
• D. Both $\left(B\right)\&\left(C\right)$

Answer 1

The correct option is D Both $\left(B\right)\&\left(C\right)$

Given:

$KE=\frac{1}{2}m{v}^2=50\text{MeV}$

$\Rightarrow \frac{1}{2}\times 1.6\times {10}^{-27}\times v^2=50\times {10}^6\times 1.6\times {10}^{-19}$

$\Rightarrow v={10}^8\text{m/s}\sim 0.3c$

The frequency of revolution is independent of the particle's speed, and it is only holds for speed that are much less than speed of light.

At very high speed, $v\sim 0.3c$, the problem must be treated with relative theory approach.

According to this theory, frequency of revolution of proton decreases compared to fixed cyclotron frequency.

Thus, the energy of the circulating proton stops increasing, and it gets out of step with the cyclotron oscillator.

$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$

$m=\frac{m_0}{\sqrt{1-[\frac{0.3c}{c}}{]}^2}$

$\therefore m>m_0$

Since frequency of the cyclotron is defined as

$f=\frac{qB}{2\pi m}$

Thus, $f$ decreases because $m$ increases.

Also, the circulating path radius at such high speed $(v\sim 0.3c)$ will be large, which makes the design of corresponding magnet to be used very expensive.

Therefore, option $\left(D\right)$ is correct.