Question

An $\alpha$ particle and proton are accelerated in cyclotron under idential conditions. Find the ratio of their cyclotron frequncy

Hint:

Cyclotron frequency is proportional to the specific charge ($q/m$) of the particle. $f \propto (q/m)$. Since an alpha particle has twice the charge but four times the mass of a proton, its specific charge is half that of a proton.

Answer 1

Step 1: Understanding the Concept:
In a cyclotron, charged particles are accelerated by an alternating electric field and kept in a spiral path by a constant magnetic field. The cyclotron frequency is the frequency of the alternating electric field, which must match the frequency of revolution of the particle to ensure continuous acceleration.

Step 2: Key Formula or Approach:
The cyclotron frequency ($f$) depends on the particle's charge ($q$), its mass ($m$), and the applied magnetic field ($B$):
\[ f = \frac{qB}{2\pi m} \]

Step 3: Detailed Explanation:
"Identical conditions" implies that the magnetic field $B$ is the same for both particles.
For a proton ($p$):
Charge $q_p = e$
Mass $m_p = m$
Cyclotron frequency $f_p = \frac{eB}{2\pi m}$
For an alpha particle ($\alpha$), which is a helium nucleus (2 protons, 2 neutrons):
Charge $q_\alpha = 2e$
Mass $m_\alpha \approx 4m$ (mass of 4 nucleons)
Cyclotron frequency $f_\alpha = \frac{(2e)B}{2\pi (4m)} = \frac{eB}{4\pi m}$
Now, find the ratio of their frequencies ($f_\alpha / f_p$):
\[ \text{Ratio} = \frac{f_\alpha}{f_p} = \frac{\frac{eB}{4\pi m}}{\frac{eB}{2\pi m}} \]
\[ \text{Ratio} = \frac{1}{4} \times \frac{2}{1} = \frac{2}{4} = \frac{1}{2} \]
The ratio of the cyclotron frequency of an $\alpha$ particle to that of a proton is $1:2$.

Step 4: Final Answer:
The ratio is $1:2$.