/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 A certain radioactive nucleus ca... [FREE SOLUTION] | 91Ó°ÊÓ

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A certain radioactive nucleus can transform to another nucleus by emitting an electron and a neutrino. (The neutrino is one of the fundamental particles of physics.) Suppose that in such a transformation, the initial nucleus is stationary, the electron and neutrino are emitted along perpendicular paths, and the magnitudes of the translational momenta are \(1.2 \times\) \(10^{-22} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) for the electron and \(6.4 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) for the neu- trino. As a result of the emissions, the new nucleus moves (recoils). (a) What is the magnitude of its translational momentum? What is the angle between its path and the path of (b) the electron (c) the neutrino?

Short Answer

Expert verified
Magnitude: \(1.36 \times 10^{-22} \text{ kg} \cdot \text{m/s}\). Angle with electron: \(28.07^\text{o}\). Angle with neutrino: \(61.93^\text{o}\).

Step by step solution

01

- Understand the problem

We need to find the magnitude of the translational momentum of the new nucleus after two particles—an electron and a neutrino—are emitted along perpendicular paths. Additionally, we need to determine the angles between the path of the new nucleus and the two particles.
02

- Identify given values

We are given that the translational momenta of the electron and the neutrino are:\[ p_e = 1.2 \times 10^{-22} \text{ kg} \times \text{m/s} \ p_u = 6.4 \times 10^{-23} \text{ kg} \times \text{m/s} \]
03

- Apply conservation of momentum

Since the initial nucleus is stationary, the vector sum of the momenta of the emitted particles and the new nucleus must be zero. Let's denote the magnitudes as follows:\[ p_{\text{nucleus}} = \text{unknown} \ p_e^2 + p_u^2 = p_{\text{nucleus}}^2 \]
04

- Calculate the magnitude of the nucleus's momentum

Using the Pythagorean theorem to find the resultant momentum: \[ p_{\text{nucleus}} = \sqrt{p_e^2 + p_u^2} = \sqrt{(1.2 \times 10^{-22})^2 + (6.4 \times 10^{-23})^2} \]Computing the values inside the square root and then taking the square root gives us the magnitude of the momentum.
05

- Solve the equation

Calculate the squared values and then take the square root:\[ p_{\text{nucleus}} = \sqrt{(1.44 \times 10^{-44}) + (4.096 \times 10^{-45})} = \sqrt{1.8496 \times 10^{-44}} = 1.36 \times 10^{-22} \text{ kg} \times \text{m/s} \]
06

- Determine the angles

(a) The magnitude of its translational momentum is\[ p_{\text{nucleus}} = 1.36 \times 10^{-22} \text{ kg} \times \text{m/s} \](b) The angle with respect to the electron's path is\[ \theta_e = \tan^{-1} \left( \frac{p_u}{p_e} \right) = \tan^{-1} \left( \frac{6.4 \times 10^{-23}}{1.2 \times 10^{-22}} \right) \]Calculating:\[ \theta_e = \tan^{-1} (0.533) = 28.07^\text{o} \](c) The angle with respect to the neutrino's path is\[ \theta_u = \tan^{-1} \left( \frac{p_e}{p_u} \right) = \tan^{-1} \left( \frac{1.2 \times 10^{-22}}{6.4 \times 10^{-23}} \right) \]Calculating:\[ \theta_u = \tan^{-1} (1.875) = 61.93^\text{o} \]

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Key Concepts

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

Radioactive Decay
Radioactive decay is a natural process through which an unstable atomic nucleus loses energy by emitting radiation. This can include a variety of particles such as alpha particles, beta particles (electrons), and neutrinos. The process transforms the original nucleus, known as the parent, into a different nucleus, known as the daughter. In our example, the nucleus emits an electron and a neutrino, leading to this transformation.
The law of conservation of momentum plays a vital role in understanding the changes in the momentum of particles emitted during the decay.
The total momentum before and after decay must remain the same since no external forces act on the system.
Understanding how momentum is conserved helps us predict the behavior of the products of radioactive decay.
Emission of Particles
When a radioactive nucleus decays, it can emit fundamental particles such as electrons and neutrinos. In this case, the electron and the neutrino are emitted along perpendicular paths.
The momentum of these particles is significant because it dictates how the remaining nucleus will move after the emission.
This perpendicular emission impacts the overall conservation of momentum and is essential for calculating the nucleus's new momentum and direction.
In our specific exercise, the electron's momentum was given as \(1.2 \times 10^{-22} \text{ kg} \times \text{m/s} \), and the neutrino's as \(6.4 \times 10^{-23} \text{ kg} \times \text{m/s} \). By using these values, we can determine the resulting momentum of the recoiling nucleus.
Translational Momentum
Translational momentum is the momentum associated with an object moving in a straight, linear path. It is given by the product of an object's mass and its velocity.
In our scenario, we are calculating the translational momentum of the new nucleus after the emission of the electron and neutrino.
Since the initial nucleus is stationary, the sum of momenta of the electron, neutrino, and new nucleus must be zero to satisfy the conservation of momentum. Using the Pythagorean theorem, which is applicable because the emission paths are perpendicular, we can find the resultant momentum:\[ p_{\text{nucleus}} = \sqrt{(1.2 \times 10^{-22})^2 + (6.4 \times 10^{-23})^2 } = 1.36 \times 10^{-22} \text{ kg} \times \text{m/s} \]
This resultant momentum helps us understand how the nucleus recoils after the particles are emitted.
Perpendicular Motion
Perpendicular motion is crucial in this problem because the electron and neutrino are emitted at right angles to each other. This orthogonal emission allows us to use the Pythagorean theorem to calculate the resultant momentum of the nucleus.
In simple terms, perpendicular motion means the direction of one particle’s movement is at a 90-degree angle relative to the other.
This relationship simplifies our calculations and lets us determine the recoil's direction and magnitude more precisely.
The angles between the paths of the new nucleus and the particles can be found using inverse tangent functions. For instance, the angle with respect to the electron’s path \( \theta_e \) is calculated as:\[ \tan^{-1} \left( \frac {p_u}{p_e} \right)= 28.07^\text{o} \]
Similarly, for the neutrino's path \( \theta_u \): \[ \tan^{-1} \left( \frac{p_e}{ p_u} \right)= 61.93^\text{o} \]
Fundamental Particles
Fundamental particles are the smallest building blocks of matter and include particles like electrons and neutrinos as discussed in our example.
These particles have no substructure, meaning they are not composed of smaller particles.
Understanding the properties and behaviors of fundamental particles helps us explain broader phenomena in physics, such as radioactive decay.
In our scenario, the electron emitted is a type of beta particle, and the neutrino is a key particle involved in weak nuclear interactions.
Knowing the characteristics and interactions of these particles is essential for solving problems involving nuclear reactions and the conservation of momentum.

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