Proton, deuteron and alpha particles of same kinetic energy are moving in circular trajectories in a

Q: Why do the particles move in a circular path?

When a charged particle enters a uniform magnetic field perpendicularly, the magnetic force acts at a right angle to its velocity. This force changes the direction of the particle without changing its speed, acting exactly like a centripetal force, which results in a circular trajectory.

Q: What exactly is a deuteron?

A deuteron is the nucleus of deuterium, which is an isotope of hydrogen. It contains one proton and one neutron. Therefore, its mass is approximately twice that of a normal hydrogen nucleus (a single proton), but its charge is exactly the same (+1e).

Q: Why is the alpha particle's radius the same as the proton's? It's much heavier!

You're right that the alpha particle is 4 times heavier, which tends to increase the radius (since $r \propto \sqrt{m}$). However, it also has twice the charge, which tends to decrease the radius (since $r \propto 1/q$). Mathematically, $\frac{\sqrt{4}}{2} = \frac{2}{2} = 1$. These two effects perfectly cancel each other out for a given kinetic energy!

Q: What if the question said they had the same momentum instead of kinetic energy?

If they had the same momentum $p$, we would use the formula $r = \frac{p}{qB}$. Since $p$ and $B$ are constant, the radius would simply be inversely proportional to the charge ($r \propto 1/q$). In that case, the proton and deuteron would have the same radius, and the alpha particle would have half that radius.

Q: Does the magnetic field do any work on the particles to keep them moving?

No, the magnetic field does zero work on the particles. Work is defined as force dotted with displacement ($W = \vec{F} \cdot d\vec{s}$). Since the magnetic force is always perpendicular to the velocity (and thus the displacement), the dot product is zero. The kinetic energy of the particles remains strictly constant.

Question

Proton, deuteron and alpha particles of same kinetic energy are moving in circular trajectories in a constant magnetic field. The radii of proton, deuteron and alpha particle are respectively $r_p$, $r_d$ and $r_\alpha$. Which one of the following relation is correct?

Options:

(A) $r_\alpha = r_p = r_d$

(B) $r_\alpha = r_p < r_d$

(C) $r_\alpha > r_d > r_p$

(D) $r_\alpha = r_d > r_p$

Sigma Percentile · Accepted Answer

### The Dance of Charged Particles in a Magnetic Field

Imagine firing three distinct subatomic particles—a proton, a deuteron, and an alpha particle—into a region with a uniform magnetic field. To make things fair, you give them all the exact same kinetic energy. As soon as they enter the magnetic field, they experience a Lorentz force that acts perpendicular to their velocity, forcing them into circular trajectories.

But will their circles be of the same size? Let's dive into the physics to find out.

### The Master Equation for Radius

When a charged particle moves perpendicularly through a magnetic field, the magnetic force provides the necessary centripetal force to keep it in a circular path. We can write this as:

$$qvB = \frac{mv^2}{r}$$

Rearranging for the radius $r$, we get:

$$r = \frac{mv}{qB} = \frac{p}{qB}$$

Here, $p$ is the momentum of the particle. However, our problem states that the particles have the same **kinetic energy** ($K$), not the same momentum. We need to bridge the gap between momentum and kinetic energy using the classic relation $K = \frac{p^2}{2m}$, which gives us $p = \sqrt{2mK}$.

Substituting this back into our radius equation, we arrive at our master formula:

$$r = \frac{\sqrt{2mK}}{qB}$$

### Finding the Proportionality

Look closely at the master formula. For all three particles, the kinetic energy $K$ and the magnetic field $B$ are identical. The number $2$ is just a constant. Therefore, the radius depends only on the mass $m$ and the charge $q$ of the specific particle. We can write this as a proportionality:

$$r \propto \frac{\sqrt{m}}{q}$$

This elegant relation tells us exactly how the radius scales with the intrinsic properties of the particles.

### Comparing the Particles

To use our proportionality, we need to know the relative masses and charges of our three contenders. Let's define the proton's mass as $m_p$ and its charge as $q_p$.

1. **Proton ($p$):** Mass $= m_p$, Charge $= q_p$
2. **Deuteron ($d$):** A deuteron is the nucleus of deuterium (1 proton + 1 neutron). So, its mass is roughly twice that of a proton, but its charge is the same. Mass $= 2m_p$, Charge $= q_p$
3. **Alpha Particle ($\alpha$):** An alpha particle is a helium nucleus (2 protons + 2 neutrons). Its mass is four times that of a proton, and its charge is twice as much. Mass $= 4m_p$, Charge $= 2q_p$

### The Final Ratio

Now, let's plug these values into our proportionality relation to find the ratio of their radii $r_p : r_d : r_\alpha$:

$$r_p : r_d : r_\alpha = \frac{\sqrt{m_p}}{q_p} : \frac{\sqrt{2m_p}}{q_p} : \frac{\sqrt{4m_p}}{2q_p}$$

We can pull out the common factors to simplify the ratio:

$$= 1 : \sqrt{2} : \frac{\sqrt{4}}{2}$$

Since $\sqrt{4} = 2$, the last term becomes $\frac{2}{2} = 1$. The ratio simplifies beautifully to:

$$= 1 : \sqrt{2} : 1$$

This result is fascinating! It tells us that the radius of the proton's path is exactly equal to the radius of the alpha particle's path ($r_p = r_\alpha$). Meanwhile, the deuteron, having a ratio of $\sqrt{2} \approx 1.414$, traces out a larger circle.

Therefore, the correct relation is **$r_\alpha = r_p < r_d$**.

Select Answer:

Visualized Solution

The Setup

Radius of Circular Path

Radius in terms of Kinetic Energy

Proportionality

Mass and Charge Relations

Calculating the Ratio

Final Conclusion

The Way Forward

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Dance of Charged Particles in a Magnetic Field

The Master Equation for Radius

Finding the Proportionality

Comparing the Particles

The Final Ratio

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