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Chapter 7: Problem 83

A radioactive nucleus is at rest when it spontaneously decays by emitting an electron and neutrino. The momentum of the electron is $8.20 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ and it is directed at right angles to that of the neutrino. The neutrino's momentum has magnitude $5.00 \times 10^{-19} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s}$ (a) In what direction does the newly formed (daughter) nucleus recoil? (b) What is its momentum?

Short Answer

Expert verified
Question: When a radioactive nucleus decays emitting an electron and a neutrino, the magnitudes of their momenta are \(8.20 \times 10^{-19}\,\text{kg} \cdot \text{m} / \text{s}\) and \(5.00 \times 10^{-19}\,\text{kg} \cdot \text{m} / \text{s}\), respectively. These vectors make a 90-degree angle. Find the direction and magnitude of the daughter nucleus' momentum vector. Answer: (a) The direction of the daughter nucleus is approximately 31.17 degrees from the electron's momentum vector. (b) The magnitude of the daughter nucleus' momentum is approximately \(9.73 \times 10^{-19} \,\text{kg} \cdot \text{m} / \text{s}\).

Step by step solution

01

Write down the given information

The momentum of the electron (m_e) is given: \(m_e = 8.20 \times 10^{-19}\,\text{kg} \cdot \text{m} / \text{s}\). The momentum of the neutrino (m_n) is given: \(m_n = 5.00 \times 10^{-19}\,\text{kg} \cdot \text{m} / \text{s}\). The angle between their directions is 90 degrees. We should find out the direction and momentum of the daughter nucleus (m_d).
02

Apply conservation of momentum

We'll apply the law of conservation of momentum, which states that the initial total momentum equals the final total momentum after the decay. So, \(m_d = m_e + m_n\). Since the momentum of the electron and neutrino are at a right angle to each other, we can find the momentum of the daughter nucleus using the Pythagorean theorem: \(m_d^2 = m_e^2 + m_n^2\) Substituting the given values: \(m_d^2 = (8.20 \times 10^{-19} \,\text{kg} \cdot \text{m} / \text{s})^2 + (5.00 \times 10^{-19} \,\text{kg} \cdot \text{m} / \text{s})^2\)
03

Calculate the momentum of the daughter nucleus

Calculate the momentum of the daughter nucleus: \(m_d = \sqrt{(8.20 \times 10^{-19})^2 + (5.00 \times 10^{-19})^2}\,\text{kg} \cdot \text{m} / \text{s}\) \(m_d = 9.7258 \times 10^{-19}\,\text{kg} \cdot \text{m} / \text{s}\) So, the momentum of the daughter nucleus is approximately \(9.73 \times 10^{-19} \,\text{kg} \cdot \text{m} / \text{s}\).
04

Determine the direction of the daughter nucleus (component method)

We can calculate the directional angle (θ) of the daughter nucleus by calculating the inverse tangent (arctangent) of the ratio of the magnitude of the two momenta (electron and neutrino): \(θ = \arctan(\frac{m_n}{m_e})\) \(θ = \arctan(\frac{5.00 \times 10^{-19}}{8.20 \times 10^{-19}}) = \arctan(\frac{5.00}{8.20})\) \(θ ≈ 31.17^{\circ}\) The direction of the daughter nucleus is approximately 31.17 degrees from the electron's momentum vector. So the answers are: (a) The direction of the newly formed daughter nucleus is 31.17 degrees from the electron's momentum vector. (b) The momentum of the daughter nucleus is approximately \(9.73 \times 10^{-19} \,\text{kg} \cdot \text{m} / \text{s}\).

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