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Chapter 9: Problem 55

(II) A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of \(9.6 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\) and \(6.2 \times 10^{-23} \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s},\) respectively. Determine the magnitude and the direction of the momentum of the second (recoiling) nucleus.

Short Answer

Expert verified
The nucleus' momentum is approximately \(11.4 \times 10^{-23}\) kg路m/s at an angle of \(32.0^\circ\).

Step by step solution

01

Identify the conservation principle

In this problem, we apply the principle of conservation of momentum. Momentum is conserved in both the x and y directions since no external forces act on the system. Initially, the total momentum is zero because the radioactive nucleus is at rest.
02

Set up the momentum equations

For the x-direction, the total initial momentum is zero, and it is shared among the recoiling nucleus, the electron, and the neutrino: \[ p_{nucleus,x} = -p_{electron} \]where the electron has a momentum of \[ p_{electron} = 9.6 \times 10^{-23} \text{ kg} \cdot \text{m/s}. \]For the y-direction:\[ p_{nucleus,y} = -p_{neutrino} \]where the neutrino has a momentum of \[ p_{neutrino} = 6.2 \times 10^{-23} \text{ kg} \cdot \text{m/s}. \]
03

Calculate the magnitude of the recoiling nucleus' momentum

Since the nucleus' momentum components balance out the electron and neutrino, the total momentum of the nucleus is given by:\[ p_{nucleus} = \sqrt{p_{nucleus,x}^2 + p_{nucleus,y}^2} \]Plug in the values:\[ p_{nucleus} = \sqrt{(9.6 \times 10^{-23})^2 + (6.2 \times 10^{-23})^2} \approx 11.4 \times 10^{-23} \text{ kg} \cdot \text{m/s}. \]
04

Determine the direction of the recoiling nucleus' momentum

The direction of the nucleus' momentum is calculated using the arctangent function:\[ \theta = \tan^{-1}\left(\frac{p_{nucleus,y}}{p_{nucleus,x}}\right) \]Plug in the values:\[ \theta = \tan^{-1}\left(\frac{6.2 \times 10^{-23}}{9.6 \times 10^{-23}}\right) \approx 32.0^\circ \]This angle is measured from the direction of the electron's momentum towards the neutrino's momentum.

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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 fascinating process where an unstable atomic nucleus loses energy by emitting radiation. This can involve the release of particles, energy, or both. During this decay process, new elements or isotopes are often formed. The decay is spontaneous, meaning it occurs without external input. In the case study exercise, a nucleus undergoes decay to produce an electron and a neutrino.
  • Types of radioactive decay: include alpha, beta, and gamma decay, each with unique characteristics.
  • Exponential decay: describes how the quantity of a radioactive substance decreases over time, often characterized by its half-life.
Understanding radioactive decay is crucial in fields such as nuclear physics and medicine, where it plays a role in both energy production and diagnostic imaging.
Momentum Vectors
Momentum vectors are vital in understanding the behavior of particles in motion. Momentum itself is a vector quantity, meaning it has both magnitude and direction, defined as the product of an object's mass and velocity \( \vec{p} = m \vec{v} \).In the given problem, the decay products have distinct momenta:
  • Electron momentum: calculated as \( p_{electron} = 9.6 \times 10^{-23} \text{ kg} \cdot \text{m/s} \).
  • Neutrino momentum: having \( p_{neutrino} = 6.2 \times 10^{-23} \text{ kg} \cdot \text{m/s} \).
The momentum conservation in both x and y directions provides crucial calculations in physics. The combined effect of these momenta illustrates vector addition, where resulting vectors represent both direction and magnitude of the recoiling nucleus's momentum.
Nuclear Recoil
Nuclear recoil occurs when a nucleus moves in response to the emission of particles. Think of it like a dancer spinning faster when extending their arms and legs inward due to conservation of angular momentum. In our scenario, the recoiling nucleus balances out the momentum lost to the electron and neutrino. Conservation of momentum dictates that the initial zero momentum from the nucleus rest state is maintained post-decay across all decay products. The momentum components of the recoiling nucleus are established by counterbalancing the emitted particles:
  • x-direction: \( p_{nucleus,x} = -p_{electron} \)
  • y-direction: \( p_{nucleus,y} = -p_{neutrino} \)
Ultimately, the direction and magnitude of the nucleus' final momentum form a vector combination of these components, calculated through techniques like Pythagorean theorem and trigonometric functions to express the angle.

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Most popular questions from this chapter

An extrasolar planet can be detected by observing the wobble it produces on the star around which it revolves. Suppose an extrasolar planet of mass \(m_{\mathrm{B}}\) revolves around its star of mass \(m_{\mathrm{A}}\) If no external force acts on this simple two-object system, then its \(\mathrm{CM}\) is stationary. Assume \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) are in circular orbits with radii \(r_{\mathrm{A}}\) and \(r_{\mathrm{B}}\) about the system's \(\mathrm{CM}\). (a) Show that \(r_{\mathrm{A}}=\frac{m_{\mathrm{B}}}{m_{\mathrm{A}}} r_{\mathrm{B}}\) (b) Now consider a Sun-like star and a single planet with the same characteristics as Jupiter. That is, \(m_{\mathrm{B}}=1.0 \times 10^{-3} \mathrm{~m}_{\mathrm{A}}\) and the planet has an orbital radius of \(8.0 \times 10^{11} \mathrm{~m} .\) Determine the radius \(r_{\mathrm{A}}\) of the star's orbit about the system's \(\mathrm{CM}\). (c) When viewed from Earth, the distant system appears to wobble over a distance of \(2 r_{\mathrm{A}}\). If astronomers are able to detect angular displacements \(\theta\) of about 1 milliarcsec \(\left(1 \operatorname{arcsec}=\frac{1}{3600}\right.\) of a degree), from what distance \(d\) (in light-years) can the star's wobble be detected \(\left(11 \mathrm{y}=9.46 \times 10^{15} \mathrm{~m}\right) ?(d)\) The star nearest to our Sun is about 4 ly away. Assuming stars are uniformly distributed throughout our region of the Milky Way Galaxy, about how many stars can this technique be applied to in the search for extrasolar planetary systems?

(III) Determine the \(\mathrm{CM}\) of a machine part that is a uniform cone of height \(h\) and radius \(R\), Fig. 9-46. [Hint: Divide the cone into an infinite number of disks of thickness \(d z,\) one of which is shown.

(III) \(\mathrm{A}\) neon atom \((m=20.0 \mathrm{u})\) makes a perfectly elastic collision with another atom at rest. After the impact, the neon atom travels away at a \(55.6^{\circ}\) angle from its original direction and the unknown atom travels away at a \(-50.0^{\circ}\) angle. What is the mass (in u) of the unknown atom? [Hint: You could use the law of sines.

(II) A 6.0 -kg object moving in the \(+x\) direction at \(5.5 \mathrm{~m} / \mathrm{s}\) collides head-on with an \(8.0-\mathrm{kg}\) object moving in the \(-x\) direction at \(4.0 \mathrm{~m} / \mathrm{s} .\) Find the final velocity of each mass if: ( \(a\) ) the objects stick together; \((b)\) the collision is elastic; \((c)\) the \(6.0-\mathrm{kg}\) object is at rest after the collision; \((d)\) the \(8.0-\mathrm{kg}\) object is at rest after the collision; \((e)\) the \(6.0-\mathrm{kg}\) object has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the \(-x\) direction after the collision. Are the results in \((c),(d),\) and \((e)\) "reasonable"? Explain.

(II) A 144-g baseball moving \(28.0 \mathrm{~m} / \mathrm{s}\) strikes a stationary 5.25-kg brick resting on small rollers so it moves without significant friction. After hitting the brick, the baseball bounces straight back, and the brick moves forward at \(1.10 \mathrm{~m} / \mathrm{s} .\) (a) What is the baseball's speed after the collision? (b) Find the total kinetic energy before and after the collision.
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