Question

The maximum kinetic energy of a proton ejected from a cyclotron with dees of radius \(70\) cm and oscillating frequency \(5\) MHz (in MeV) is \[ (\text{Mass of proton}=1.67\times10^{-27}\text{ kg}) \]

• A. \(2.53\)
• B. \(3.87\)
• C. \(1.46\)
• D. \(4.84\)

Hint:

For a cyclotron, \[ \boxed{ K_{\max} = \frac12m(2\pi fr)^2. } \] After obtaining energy in joules, convert it to MeV using \[ \boxed{ 1\text{ MeV}=1.6\times10^{-13}\text{ J}. } \]

Answer 1

The Correct Option is A

Step 1: Use the cyclotron relation. The maximum kinetic energy of a charged particle in a cyclotron is \[ K_{\max} = \frac12m(2\pi fr)^2, \] where \[ m=1.67\times10^{-27}\text{ kg}, \] \[ f=5\times10^6\text{ Hz}, \] and \[ r=70\text{ cm}=0.7\text{ m}. \]

Step 2: Calculate the kinetic energy. Substituting, \[ K_{\max} = \frac12(1.67\times10^{-27}) \left(2\pi\times5\times10^6\times0.7\right)^2. \] This gives \[ K_{\max} \approx 4.05\times10^{-13}\text{ J}. \] Since \[ 1\text{ MeV} = 1.6\times10^{-13}\text{ J}, \] \[ K_{\max} = \frac{4.05\times10^{-13}} {1.6\times10^{-13}} \approx2.53\text{ MeV}. \]

Step 3: Write the answer. Hence, \[ \boxed{2.53\text{ MeV}}. \] Thus, \[ \boxed{(A)} \] is the correct answer.