A proton, a deutron and an alpha-particle having the same kinetic energy are moving in circular traj

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

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts perpendicular to the velocity, providing the centripetal force required for circular motion.

Q: How is momentum related to kinetic energy?

Kinetic energy is $K = \frac{1}{2}mv^2$. Multiplying both sides by $m$ gives $mK = \frac{1}{2}m^2v^2 = \frac{1}{2}p^2$. Solving for $p$ gives $p = \sqrt{2mK}$.

Q: What is a deuteron?

A deuteron is the nucleus of deuterium, an isotope of hydrogen. It contains one proton and one neutron, so its mass is roughly twice that of a proton, but its charge is the same ($+e$).

Q: Why do the proton and alpha particle have the same radius?

Even though the alpha particle is four times heavier, its momentum is only twice as large (since $p \propto \sqrt{m}$). Because its charge is also twice as large, the factor of 2 cancels out in the ratio $\frac{\sqrt{m}}{q}$, resulting in the same radius as the proton.

Question

A proton, a deutron and an $\alpha$-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If $r_p, r_d$ and $r_\alpha$ denote, respectively the radii of the trajectories of these particles, then

Options:

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

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

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

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

Sigma Percentile · Accepted Answer

### The Dance of Particles in a Magnetic Field

Imagine a proton, a deuteron, and an alpha particle, all injected into a uniform magnetic field with the exact same kinetic energy. They will all trace circular paths, but with different radii. Let's find out how these radii compare.

When a charged particle moves perpendicularly through a magnetic field, it experiences a magnetic Lorentz force that provides the necessary centripetal force for circular motion. The radius of this circular path is given by the well-known formula:

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

where $m$ is the mass, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength.

### Connecting Radius to Kinetic Energy

The problem doesn't give us the velocity of the particles; instead, it tells us they have the same kinetic energy, $K$. We need to express our radius formula in terms of $K$.

We know that momentum $p = mv$ can be related to kinetic energy through the equation $p = \sqrt{2mK}$. Substituting this into our radius formula, we get a new expression:

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

Now, look closely at this equation. The kinetic energy $K$ and the magnetic field $B$ are the same for all three particles. The number $2$ is just a constant. This means the radius $r$ is directly proportional to the square root of the mass divided by the charge:

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

### Comparing the Particles

Let's list the relative masses and charges of our three particles. Let the proton have mass $m$ and charge $e$.

A deuteron is the nucleus of deuterium, consisting of one proton and one neutron. Therefore, it has a mass of $2m$ and a charge of $e$.

An alpha particle is a helium nucleus, consisting of two protons and two neutrons. Thus, it has a mass of $4m$ and a charge of $2e$.

Now, let's plug these values into our proportionality relation to find the ratio of their radii:

$$r_p : r_d : r_\alpha = \frac{\sqrt{m}}{e} : \frac{\sqrt{2m}}{e} : \frac{\sqrt{4m}}{2e}$$

Simplifying this ratio, we get:

$$r_p : r_d : r_\alpha = 1 : \sqrt{2} : \frac{2}{2}$$

$$r_p : r_d : r_\alpha = 1 : \sqrt{2} : 1$$

### The Final Verdict

Comparing these values, we see that the proton and the alpha particle have the exact same radius, which is $1$ in our relative units. The deuteron has a larger radius of $\sqrt{2}$, which is approximately $1.414$.

Therefore, the relationship between their radii is:

**$$r_\alpha = r_p < r_d$$**

Select Answer:

Visualized Solution

$r_{p}, r_{d}, r_{α}$

$r = \frac{m v}{q B}$

$p = 2 m K$

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

Mass and Charge

Ratio of Radii

$r_{α} = r_{p} < r_{d}$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Dance of Particles in a Magnetic Field

Connecting Radius to Kinetic Energy

Comparing the Particles

The Final Verdict

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