Problem 75 - Fundamentals of Physics

Question

A proton, a deuteron \((q=+e, m=2.0 \mathrm{u}),\) and an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) all having the same kinetic energy enter a region of uniform magnetic field \(\vec{B},\) moving perpendicular to \(\vec{B}\) What is the ratio of (a) the radius \(r_{d}\) of the deuteron path to the radius \(r_{p}\) of the proton path and (b) the radius \(r_{a}\) of the alpha particle path to \(r_{p} ?\)

Step-by-Step Solution

Answer

\( r_d : r_p = 1.41:1 \) and \( r_a : r_p = 1.41:1 \).

Step 1: Understand the Relationship between Radius and Mass, Charge, and Velocity

The radius of the path of a charged particle moving perpendicular to a magnetic field is given by the equation \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.

Step 2: Equate Kinetic Energies of Particles

Since the proton, deuteron, and alpha particle have the same kinetic energy, we can write their kinetic energies as: \( \frac{1}{2}m_pv_p^2 = \frac{1}{2}m_dv_d^2 = \frac{1}{2}m_av_a^2 \). From this, we get \( m_pv_p^2 = m_dv_d^2 = m_av_a^2 \).

Step 3: Derive Velocity in Terms of Mass and Kinetic Energy

Since kinetic energy is the same for all, solve for velocity: \( v = \sqrt{\frac{2K}{m}} \), where \( K \) is the kinetic energy. So, \( v_p = \sqrt{\frac{2K}{m_p}} \), \( v_d = \sqrt{\frac{2K}{m_d}} \), \( v_a = \sqrt{\frac{2K}{m_a}} \).

Step 4: Substitute Velocities into Radius Formula

Substitute the velocities into the radius formula: \( r = \frac{m}{qB}v \), gives \( r_p = \frac{m_p}{q_pB}v_p = \frac{m_p}{q_pB}\sqrt{\frac{2K}{m_p}} \). Simplify, \( r_p = \frac{1}{q_pB}\sqrt{2Km_p} \).

Step 5: Calculate Radius Ratios

For the deuteron: \( r_d = \frac{1}{q_dB}\sqrt{2Km_d} \). The radius ratio \( \frac{r_d}{r_p} = \frac{\sqrt{m_d}/q_d}{\sqrt{m_p}/q_p} \). For the alpha particle: \( r_a = \frac{1}{q_aB}\sqrt{2Km_a} \). The radius ratio \( \frac{r_a}{r_p} = \frac{\sqrt{m_a}/q_a}{\sqrt{m_p}/q_p} \).

Step 6: Substitute Masses and Charges

Substitute the given values: for deuteron \( m_d = 2u, q_d = +e \), for alpha \( m_a = 4u, q_a = +2e \), and for proton \( m_p = 1u, q_p = +e \). Therefore, \( \frac{r_d}{r_p} = \sqrt{\frac{2}{1}} \) and \( \frac{r_a}{r_p} = \sqrt{\frac{4}{2}} \).

Step 7: Simplify the Ratios

Simplify \( \frac{r_d}{r_p} = \sqrt{2} \approx 1.41 \) and \( \frac{r_a}{r_p} = \sqrt{2} = 1.41 \).

Key Concepts

Proton Path Radius

When a charged particle, like a proton, moves through a magnetic field perpendicular to its velocity, it follows a circular path. This is because the magnetic force acts as a centripetal force, pulling the charged particle into a curved trajectory. The radius of this path, termed as the proton path radius, depends on several factors:

• Mass of the proton (\( m_p \)).
• Velocity of the proton (\( v_p \)).
• Charge of the proton (\( q_p \)).
• Strength of the magnetic field (\( B \)).

The formula that gives us the radius is \( r = \frac{mv}{qB} \).
For a proton, this becomes \( r_p = \frac{m_p}{q_p B} \sqrt{\frac{2K}{m_p}} \), where \( K \) is the kinetic energy. In a uniform magnetic field, a proton moves in a path whose radius helps us understand its motion under the influence of a magnetic field.

Charge-to-Mass Ratio

The charge-to-mass ratio is crucial in determining how a charged particle behaves in a magnetic field. This ratio defines how much charge a particle has relative to its mass, influencing the radius of its path. Particles with higher charge-to-mass ratios experience stronger magnetic forces, causing tighter curves in magnetic fields.
Here's the role it plays:

• A higher charge (\( q \)) or a lower mass (\( m \)) results in a larger charge-to-mass ratio.
• This leads to more pronounced bending and larger acceleration in a magnetic field.

In the context of the given problem:

• The proton has \( q_p = +e \) and \( m_p = 1u \)
• The deuteron has \( q_d = +e \) and \( m_d = 2u \)
• The alpha particle has \( q_a = +2e \) and \( m_a = 4u \)

Thus, comparing their charge-to-mass ratios helps in calculating the path radius ratios like \( \frac{r_d}{r_p} \) and \( \frac{r_a}{r_p} \).
This comparison is pivotal to understanding how different particles traverse through identical magnetic fields.

Kinetic Energy Equality

Understanding kinetic energy equality is vital to solve problems involving multiple particles moving in magnetic fields. By setting the kinetic energies of the proton, deuteron, and alpha particles equal, we assume they have equivalent capacities for motion. This simplifies calculations by tying it all back to their respective velocities. The shared kinetic energy \( K \) is given by the equation \( \frac{1}{2} mv^2 \).
Since \( K \) is the same for all particles, it allows us to express their velocities:

• For the proton: \( v_p = \sqrt{\frac{2K}{m_p}} \)
• For the deuteron: \( v_d = \sqrt{\frac{2K}{m_d}} \)
• For the alpha particle: \( v_a = \sqrt{\frac{2K}{m_a}} \)

This relationship provides a foundation for deriving the path radius equation, \( r = \frac{mv}{qB} \).
Thus, understanding kinetic energy equality helps determine each particle's path radius by tying together velocity, charge, mass, and its motion in the magnetic field.