Question

In a cyclotron, for the same values of B and R, what is the ratio of the maximum kinetic energies of \(\alpha\) particles and deuteron? {Mass of deuteron \(=2m_{p}\), charge on deuteron \(=+e\)}

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

At the exit radius the kinetic energy depends on charge squared divided by mass for fixed field and radius. Compare this quantity for the two particles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
At the exit radius, \(r=\dfrac{mv}{qB}=R\), so \(v=\dfrac{qBR}{m}\) and \(K=\dfrac{q^{2}B^{2}R^{2}}{2m}\). For fixed B and R, \(K\propto\dfrac{q^{2}}{m}\).

Step 2: Alpha particle
\(q=2e\) and \(m=4m_{p}\), so \(\dfrac{q^{2}}{m}=\dfrac{4e^{2}}{4m_{p}}=\dfrac{e^{2}}{m_{p}}\).

Step 3: Deuteron and ratio
\(q=e\) and \(m=2m_{p}\), so \(\dfrac{q^{2}}{m}=\dfrac{e^{2}}{2m_{p}}\).
The ratio is \(\dfrac{e^{2}/m_{p}}{e^{2}/2m_{p}}=2\).

Final Answer:
The ratio of maximum kinetic energies is 2. \[ \boxed{\text{(B) }2} \]