A proton and an alpha-particle, having kinetic energies K_p and K_alpha, respectively, enter into a

Q: Why do we use momentum instead of velocity to solve this?

Because kinetic energy is directly related to momentum via $K = \frac{p^2}{2m}$. Finding the momentum ratio first makes the calculation much smoother and less prone to algebraic errors than finding the velocity ratio and squaring it later.

Q: What are the exact charge and mass of an alpha particle compared to a proton?

An alpha particle is a helium nucleus, meaning it contains 2 protons and 2 neutrons. Therefore, its charge is $+2e$ (twice that of a proton) and its mass is approximately $4m_p$ (four times that of a proton).

Q: Why does the magnetic field $B$ cancel out in the ratio?

Both particles enter the *same* uniform magnetic field. When we take the ratio of their radii, the $B$ term is present in the denominator of both expressions, so it cancels out perfectly.

Q: What if they entered the magnetic field at an angle other than 90 degrees?

If they entered at an angle $\theta$, their paths would be helical rather than purely circular. The radius of the helix would depend on the perpendicular component of velocity, $v \sin\theta$. However, the problem explicitly specifies they enter at right angles ($\theta = 90^\circ$).

Q: Can I solve this by finding the velocity ratio first?

Yes, absolutely! You can find $\frac{v_p}{v_\alpha}$ using the radius formula $r = \frac{mv}{qB}$, and then use $K = \frac{1}{2}mv^2$. It will give you the exact same result, but using the momentum shortcut is often quicker for competitive exams.

Question

A proton and an $\alpha$-particle, having kinetic energies $K_p$ and $K_\alpha$, respectively, 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

Options:

(A) $1:8$

(B) $8:1$

(C) $1:4$

(D) $4:1$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are observing a microscopic race track. A proton and an $\alpha$-particle are shot into a uniform magnetic field at right angles to the field lines. Because the magnetic force acts perpendicularly to their velocity, it acts as a centripetal force, causing both particles to trace out circular paths.

The problem gives us a fascinating piece of data: the radius of the proton's trajectory is exactly twice that of the $\alpha$-particle. Our mission is to use this geometric clue to uncover the ratio of their kinetic energies.

### The Master Equation for Radius
To bridge the gap between the physical path and the particle's properties, we need the master equation for the radius of a charged particle in a magnetic field. By equating the magnetic Lorentz force to the centripetal force ($qvB = \frac{mv^2}{r}$), we get:

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

Since mass times velocity ($mv$) is momentum ($p$), we can elegantly rewrite this as:

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

This form is incredibly powerful because it directly links the radius to momentum, which is a stepping stone to kinetic energy.

### Unlocking the Momentum Ratio
Let's set up a ratio for the two particles. Since they enter the *same* magnetic field, $B$ is constant and will beautifully cancel out:

$$\frac{r_p}{r_\alpha} = \frac{p_p}{p_\alpha} 	imes \frac{q_\alpha}{q_p}$$

We are given that $\frac{r_p}{r_\alpha} = \frac{2}{1}$. We also know our fundamental particles: an $\alpha$-particle (a helium nucleus) has a charge of $+2e$, while a proton has a charge of $+e$. Therefore, $\frac{q_\alpha}{q_p} = 2$.

Substituting these into our ratio:

$$\frac{2}{1} = \frac{p_p}{p_\alpha} 	imes 2$$

$$\Rightarrow \frac{p_p}{p_\alpha} = 1$$

This is a thrilling revelation! Despite having different masses, charges, and path radii, the proton and the $\alpha$-particle entered the magnetic field with the **exact same momentum**.

### The Final Leap to Kinetic Energy
Now, we need to translate this momentum ratio into a kinetic energy ratio. The classic relationship between kinetic energy ($K$) and momentum ($p$) is:

$$K = \frac{p^2}{2m}$$

Let's construct the final ratio:

$$\frac{K_p}{K_\alpha} = \frac{\frac{p_p^2}{2m_p}}{\frac{p_\alpha^2}{2m_\alpha}} = \left(\frac{p_p}{p_\alpha}ight)^2 	imes \frac{m_\alpha}{m_p}$$

We already know the momentum ratio is $1$. What about the mass ratio? An $\alpha$-particle consists of two protons and two neutrons, making it approximately four times as massive as a single proton ($m_\alpha = 4m_p$).

Plugging it all in:

$$\frac{K_p}{K_\alpha} = (1)^2 	imes \frac{4m_p}{m_p} = \frac{4}{1}$$

The kinetic energy of the proton is four times that of the $\alpha$-particle. The elegance of breaking the problem down via momentum makes the final calculation a breeze!

Select Answer:

Visualized Solution

Visualizing the Trajectories

Radius of a Charged Particle

Setting up the Ratio

Calculating Momentum Ratio

Kinetic Energy and Momentum

Ratio of Kinetic Energies

Final Calculation

The Way Forward

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup Imagine you are observing a microscopic race track

The Master Equation for Radius To bridge the gap between the physical path and the particle's properties, we need the master equation for the radius of a charged particle in a magnetic field

Unlocking the Momentum Ratio Let's set up a ratio for the two particles

The Final Leap to Kinetic Energy Now, we need to translate this momentum ratio into a kinetic energy ratio

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