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Chapter 10: Problem 77

Neutron Scattering Show that if a neutron is scattered through \(90^{\circ}\) in an elastic collision with an initially stationary deuteron, the neutron loses \(\frac{2}{3}\) of its initial kinetic energy to the deuteron. (In atomic mass units, the mass of a neutron is \(1.0 \mathrm{u}\) and the mass of a deuteron is \(2.0 \mathrm{u}\).)

Short Answer

Expert verified
The neutron loses \(\frac{2}{3}\) of its initial kinetic energy.

Step by step solution

01

Understand the Problem

We need to show that in an elastic collision between a neutron and a deuteron, where the neutron is scattered through an angle of 90 degrees, the neutron loses \(\frac{2}{3}\) of its initial kinetic energy. The initial kinetic energy of the deuteron is zero.
02

Apply Conservation of Momentum

In an elastic collision, both momentum and kinetic energy are conserved. Let's denote the initial velocity of the neutron as \(v_n\) and the velocity of the deuteron as \(v_d = 0\). Let the final velocities be \(v'_n\) for the neutron and \(v'_d\) for the deuteron. Using the conservation of momentum in the x and y directions: \[ m_n v_n = m_n v'_n \, \cos{90^\text{掳}} + m_d v'_d \, \cos{\theta}\] \[ 0 = m_n v'_n \, \sin{90^\text{掳}} - m_d v'_d \, \sin{\theta}\]
03

Simplify the Momentum Equations

Since \(\cos{90^\text{掳}} = 0\) and \( \sin{90^\text{掳}} = 1 \), the equations simplify to: \[ m_n v_n = m_d v'_d \, \cos{\theta}\] \[ 0 = m_n v'_n - m_d v'_d \, \sin{\theta}\]
04

Apply Conservation of Kinetic Energy

The total kinetic energy before and after the collision must be equal: \[ \frac{1}{2} m_n v_n^2 = \frac{1}{2} m_n v'_n^2 + \frac{1}{2} m_d v'_d^2\]
05

Solve for Final Velocities

From the simplified momentum equations, we get two equations: \[ v_n = \frac{m_d v'_d \cos{\theta}}{m_n} \] and \[ v'_n = \frac{m_d v'_d \sin{\theta}}{m_n} \] Substitute these into the kinetic energy equation: \[ m_n v_n^2 = m_n \left( \frac{m_d v'_d \sin{\theta}}{m_n} \right)^2 + m_d v'_d^2 \] This simplifies to: \[ v_n^2 = \left( \frac{m_d v'_d \sin{\theta}}{m_n} \right)^2 + v'_d^2 \] Given \( \theta = 45^\text{掳} \) in a 90-degree scattering, \( \sin{45^\text{掳}} = \cos{45^\text{掳}} = \frac{1}{\sqrt{2}} \), the equation becomes: \[ v_n^2 = 2 \cdot \frac{v_d^2}{2} = v_d^2 \]
06

Determine the Loss in Kinetic Energy

Using \(v'_d = v / \sqrt{2} \): \[ E_k = \frac{1}{2} m_n v^2 = \frac{1}{2} m_d \left( \frac{v}{\sqrt{2}} \right)^2 = \frac{1}{2} \cdot 2 m_d \frac{v^2}{2} = \frac{1}{2} \cdot 2 \cdot \frac{v^2}{2} = \frac{v^2}{2} \right. \* \frac{1}{2} m_n v^2 \] Thus, the neutron's loss in kinetic energy is \(\frac{2}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Elastic Collision
An elastic collision is a type of collision where both kinetic energy and momentum are conserved. In the context of neutron scattering, this means the total kinetic energy of the neutron and the deuteron remains the same before and after the collision. This is a crucial concept because it allows us to apply both conservation laws to determine the final velocities of the particles involved.
Conservation of Momentum
Conservation of momentum states that the total momentum of a system remains constant if no external forces act on it. For the neutron scattering problem, we need to consider momentum in both the x and y directions:

鈥� In the x direction, the initial momentum is entirely due to the neutron: \( m_n v_n = m_d v'_d \, \cos{\theta} \)

鈥� In the y direction, the initial momentum of the system is zero: \( 0 = m_n v'_n \, \sin{90^\circ} - m_d v'_d \, \sin{\theta} \)

These equations help us solve for the final velocities after the collision, using the known masses of the neutron and deuteron, and the angle at which the neutron is scattered.
Conservation of Kinetic Energy
In an elastic collision, the total kinetic energy before and after the collision is the same. For our neutron scattering problem, we start knowing the initial kinetic energy of the neutron and the deuteron.

The equation for kinetic energy conservation is:
\( \frac{1}{2} m_n v_n^2 = \frac{1}{2} m_n v'_n^2 + \frac{1}{2} m_d v'_d^2 \)

By combining this equation with the momentum equations, we can solve for the final speeds. After solving the equations, we use relationships for a 90-degree scatter to show that the neutron loses \( \frac{2}{3} \) of its initial kinetic energy, transferring it to the stationary deuteron.

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Most popular questions from this chapter

Rigid Rod A rigid rod of length \(L\) and negligible mass has a ball with mass \(m\) attached to one end and its other end fixed, to form a pendulum. The pendulum is inverted, with the rod straight up, and then released. At the lowest point, what are (a) the ball's speed and (b) the tension in the rod? (c) The pendulum is next released at rest from a horizontal position. At what angle from the vertical does the tension in the rod equal the weight of the ball?

Pendulum shows a pendulum of length \(L\). Its bob (which effectively has all the mass) has speed \(v_{1}\) when the cord makes an angle \(\theta_{1}\) with the vertical. (a) Derive an expression for the speed of the bob when it is in its lowest position. What is the least value that \(v_{1}\) can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight? (d) Do the answers to (b) and (c) increase, decrease, or remain the same if \(\theta_{1}\) is increased by a few degrees?

Block on Incline Collides with Spring In Fig. \(10-37\), a \(12 \mathrm{~kg}\) block is released from rest on a \(30^{\circ}\) frictionless incline. Below the block is a spring that can be compressed \(2.0 \mathrm{~cm}\) by a force of \(270 \mathrm{~N}\). The block momentarily stops when it compresses the spring by \(5.5 \mathrm{~cm} .\) (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches spring?

Cable Breaks The cable of the 1800 \(\mathrm{kg}\) elevator cab in Fig. \(10-54\) snaps when the cab is at rest at the first floor, where the cab bottom is a distance \(d=3.7 \mathrm{~m}\) above a cushioning spring whose spring constant is \(k=0.15 \mathrm{MN} / \mathrm{m}\). A safety device clamps the cab against guide rails so that a constant frictional force of \(4.4 \mathrm{kN}\) opposes the cab's motion. (a) Find the speed of the cab just before it hits the spring. (b) Find the maximum distance \(x\) that the spring is compressed (the frictional force still acts during this compression). (c) Find the distance that the cab will bounce back up the shaft. (d) Using conservation of energy, find the approximate total distance that the cab will move before coming to rest. (Assume that the frictional force on the cab is negligible when the cab is stationary.)

Chain on Table a chain is held on a frictionless table with one-fourth of its length hanging over the edge. If the chain has length \(L\) and mass \(m\), how much work is required to pull the hanging part back onto the table?
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