A proton, an electron and a helium nucleus, have the same energy. They are in circular orbits in a p

Q: Why do electrons and protons curve in opposite directions in the diagram?

Because they have opposite charges! The Lorentz force is given by $\vec{F} = q(\vec{v} \times \vec{B})$. If a positive charge (proton) is pushed upwards, a negative charge (electron) moving in the same direction will be pushed downwards.

Q: My textbook says the answer is (c) $r_e < r_p < r_{He}$. Why is that wrong?

That is a known error in some books. The flawed solution assumes that the particles have the same *velocity* $v$, which would make $r_{He} = \frac{4m_p v}{2eB} = 2r_p$. However, the question explicitly states they have the same *energy*. When you correctly substitute $v = \sqrt{\frac{2K}{m}}$, the radii of the proton and alpha particle become exactly equal.

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

If they had the same momentum $p$, then the radius formula $r = \frac{p}{qB}$ shows that radius only depends inversely on charge $q$. Since the electron and proton have the same charge magnitude $e$, they would have the same radius ($r_e = r_p$). The alpha particle has double the charge ($2e$), so its radius would be half as large ($r_{He} = r_p / 2$).

Q: What if they were accelerated through the same potential difference V?

If accelerated through the same voltage $V$, their kinetic energy becomes $K = qV$. Substituting this into our radius formula gives $r = \frac{\sqrt{2m(qV)}}{qB} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$. In this case, the radius is proportional to $\sqrt{\frac{m}{q}}$.

Q: Is a helium nucleus exactly the same as an alpha particle?

Yes! A helium atom stripped of its two electrons is just a bare nucleus containing 2 protons and 2 neutrons. In physics, this is exactly what an alpha particle is.

Question

A proton, an electron and a helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let $r_p$, $r_e$ and $r_{He}$ be their respective radii, then Options: (A) $r_e < r_p = r_{He}$ (B) $r_e > r_p = r_{He}$ (C) $r_e < r_p < r_{He}$ (D) $r_e > r_p > r_{He}$

Sigma Percentile · Accepted Answer

### The Trap of "Same Energy"

When charged particles enter a uniform magnetic field perpendicularly, they experience a Lorentz force that acts as a centripetal force, forcing them into circular orbits. The radius of this orbit is a classic physics derivation:

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

Here is where many students (and even some textbook authors!) fall into a trap. The question explicitly states that the particles have the **same energy** (kinetic energy, $K$). It does *not* say they have the same velocity. If you assume they have the same velocity, you will get the wrong answer.

### Expressing Radius in Terms of Kinetic Energy

To compare the radii fairly, we must express the momentum $p = mv$ in terms of the given constant, kinetic energy $K$. We know the relationship between kinetic energy and momentum is:

$$K = \frac{p^2}{2m} \implies p = \sqrt{2mK}$$

Substituting this back into our radius formula, we get the master equation for this problem:

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

Since the kinetic energy $K$ and the magnetic field $B$ are identical for all three particles, we can strip away the constants and focus purely on the proportionality:

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

### The Showdown: Electron vs. Proton vs. Alpha Particle

Now, let's evaluate this ratio for our three contenders:

**1. The Electron ($e^-$):**
It has a mass $m_e$ and a charge magnitude of $e$. Its radius is proportional to $\frac{\sqrt{m_e}}{e}$.

**2. The Proton ($p^+$):**
It has a mass $m_p$ and a charge magnitude of $e$. Since the proton is roughly 1836 times more massive than the electron ($m_p \gg m_e$), its numerator is much larger. Therefore, the proton will trace a much wider circle than the electron:

$$r_e < r_p$$

**3. The Helium Nucleus (Alpha Particle, $He^{2+}$):**
A helium nucleus consists of 2 protons and 2 neutrons. This gives it a mass of approximately $4m_p$ and a charge of $+2e$. Let's plug these into our proportionality:

$$r_{He} \propto \frac{\sqrt{4m_p}}{2e} = \frac{2\sqrt{m_p}}{2e} = \frac{\sqrt{m_p}}{e}$$

Look at that beautiful cancellation! The factor of 2 from the square root of the mass perfectly cancels the factor of 2 from the charge. The resulting ratio is identical to that of the proton. Therefore:

$$r_{He} = r_p$$

### The Final Verdict

Combining our results, we find that the electron has the smallest radius, while the proton and the helium nucleus share the exact same larger radius.

**Final Relation:** $r_e < r_p = r_{He}$

*Note: Some reference materials incorrectly state that $r_{He} > r_p$ by mistakenly assuming the particles have the same velocity. Always trust the rigorous mathematical substitution!*

Animated Solution for Physics - Magnetic Effects of Current: A proton, an electron and a helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicular to the plane. Let $r_{p}$ , $r_{e}$ and $r_{H e}$ be their respective radii, then

Select Answer:

Visualized Solution

Lorentz Force & Circular Path

Kinetic Energy Substitution

Radius in terms of Energy

Electron vs Proton

Proton vs Alpha Particle

Alpha Particle Radius

Final Conclusion

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Trap of "Same Energy"

Expressing Radius in Terms of Kinetic Energy

The Showdown

The Final Verdict

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