Question

Ionized hydrogen atoms and $\alpha$-particles with same momenta enters perpendicular to a constant magnetic field. B. The ratio of their radii of their paths $r_H : r_\alpha$ will be :

• A. 4 : 1
• B. 1 : 4
• C. 2 : 1
• D. 1 : 2

Answer 1

The Correct Option is C

 Let's solve this problem step-by-step by understanding the physics involved when ionized hydrogen atoms and $\alpha$-particles enter a magnetic field.

Concept: When a charged particle enters a magnetic field perpendicular to its velocity, it experiences a centripetal force, which makes it move in a circular path. The radius $r$ of this path is given by the formula:

\(r = \frac{p}{qB}\)

where:

• \(p\) is the momentum of the particle,
• \(q\) is the charge of the particle,
• \(B\) is the magnetic field.

 

Here, both the ionized hydrogen atoms and the $\alpha$-particles have the same momentum.

Charge Consideration:

• An ionized hydrogen atom (essentially a proton) has charge \(q_H = +e\).
• An $\alpha$-particle is a helium nucleus consisting of 2 protons and 2 neutrons, so its charge is \(q_\alpha = 2e\).

Application of the Formula:

• For ionized hydrogen, the radius is \(r_H = \frac{p}{eB}\).
• For $\alpha$-particles, the radius is \(r_\alpha = \frac{p}{2eB}\).

Ratio of Radii:

When we take the ratio of $r_H$ to $r_\alpha$, we get:

\(\frac{r_H}{r_\alpha} = \frac{p}{eB} \div \frac{p}{2eB} = \frac{p}{eB} \times \frac{2eB}{p} = 2\)

Thus, the ratio of the radii of their paths is \(2:1\).

Conclusion: The correct answer is 2:1, meaning the path radius of the hydrogen atoms is twice that of the $\alpha$-particles.