Question

A proton and an \(\alpha\)-particle enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of \(\alpha\)-particle is 2 : 1. The ratio of \(K_p : K_{\alpha}\) is:

• A. 1 : 4
• B. 4 : 1
• C. 8 : 1
• D. 1 : 8

Hint:

For these problems, write the ratio formula first: \(r \propto \sqrt{mK}/q\). Squaring both sides makes solving for \(K\) much easier.

Answer 1

The Correct Option is B

Step 1: The formula for radius \(r\) in a magnetic field is \(r = \frac{\sqrt{2mK}}{qB}\).
Step 2: Express Kinetic Energy \(K\): \(K \propto \frac{q^2 r^2}{m}\) (since \(B\) is constant).
Step 3: Let proton have \(m_p = m, q_p = e\) and \(\alpha\)-particle have \(m_{\alpha} = 4m, q_{\alpha} = 2e\).
Step 4: Calculate the ratio: \[\frac{K_p}{K_{\alpha}} = \left(\frac{q_p}{q_{\alpha}}\right)^2 \times \left(\frac{r_p}{r_{\alpha}}\right)^2 \times \left(\frac{m_{\alpha}}{m_p}\right)\] \[\frac{K_p}{K_{\alpha}} = \left(\frac{e}{2e}\right)^2 \times \left(\frac{2}{1}\right)^2 \times \left(\frac{4m}{m}\right) = \frac{1}{4} \times 4 \times 4 = 4\] The ratio is 4 : 1.