91Ó°ÊÓ

A gamma ray (an energetic photon) falls on a nucleus of initial mass \(m\), initially at rest. The energy \(E_{\mathrm{p}}\) of the incoming gamma ray matches the energy separation between the lowest energy of the nucleus and its first excited state, so the incident photon is absorbed. We want to know the mass \(m^{*}\) of the excited nucleus. (see Fig. \(38-13 .\) ) (a) Show that the conservation of energy and momentum equations are, in an obvious notation: and $$ \begin{array}{c} E_{\mathrm{p}}+m c^{2}=E_{m^{*}} \\ \frac{E_{\mathrm{p}}}{c}=p_{m^{*}}=\frac{\left(E_{m^{*}}^{2}-m^{* 2} c^{4}\right)^{1 / 2}}{c} . \end{array} $$ (b) Combine the two conservation equations to find an expression for \(m^{*}\) as a function of \(E_{\mathrm{p}}, m\), and \(c\). (c) Show that for very small values of \(E_{\mathrm{p}}\) the limiting result is \(m^{*}=m .\) Explain why this limiting result is reasonable.

Step by step solution

01

Conservation of Energy Equation

Start by applying conservation of energy. The initial energy is the sum of the energy of the gamma ray and the rest energy of the nucleus. After absorption, the total energy is the rest energy of the excited nucleus. This gives us the equation: \[ E_{\text{p}} + mc^2 = E_{m^*} \]

02

Conservation of Momentum Equation

Next, apply conservation of momentum. The initial momentum is just the momentum of the gamma ray, which corresponds to its energy divided by the speed of light, c. After absorption, this momentum is the momentum of the excited nucleus. Thus, we have: \[ \frac{E_{\text{p}}}{c} = p_{m^*} = \frac{\left(E_{m^*}^2 - m^{*2}c^4\right)^{1/2}}{c} \]

03

Express Energy of Excited Nucleus

Rearrange the conservation of momentum equation to express the energy of the excited nucleus in terms of its rest mass and momentum: \[ E_{m^*} = \left(m^{*2}c^4 + \left(\frac{E_{\text{p}}}{c}\right)^2c^2\right)^{1/2} = \left(m^{*2}c^4 + E_{\text{p}}^2\right)^{1/2} \]

04

Substitute Energy Expression

Substitute this expression for \(E_{m^*}\) back into the conservation of energy equation: \[ E_{\text{p}} + mc^2 = \left(m^{*2}c^4 + E_{\text{p}}^2\right)^{1/2} \]

05

Isolate for Excited Mass

Square both sides of the equation to eliminate the square root: \[ \left(E_{\text{p}} + mc^2\right)^2 = m^{*2}c^4 + E_{\text{p}}^2 \] Then, solve for \(m^*\): \[ m^{*2}c^4 = \left(E_{\text{p}} + mc^2\right)^2 - E_{\text{p}}^2 \]\[ m^{*2}c^4 = E_{\text{p}}^2 + 2E_{\text{p}}mc^2 + m^2c^4 - E_{\text{p}}^2 \] \[ m^{*2}c^4 = 2E_{\text{p}}mc^2 + m^2c^4 \] Thus, \[ m^{*} = \left(2\frac{E_{\text{p}}}{c^2}mc^2 + m^2\right)^{1/2} \] Simplifying further: \[ m^{*} = \left(2\frac{E_{\text{p}}}{c^2}m + m^2\right)^{1/2} \] .

06

Limiting Case for Small Energy

When \(E_{\text{p}}\) is very small compared to \(mc^2\), the term \(2\frac{E_{\text{p}}}{c^2}m\) is negligible. Therefore, \[ m^{*} = m \] This limiting result is reasonable because if the energy of the gamma photon is very small, its contribution to the mass of the nucleus is negligible, causing no significant increase in the mass.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Conservation of Energy

In nuclear physics, the law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. This principle is crucial when solving problems involving nuclear reactions, such as the scenario where a gamma ray is absorbed by a nucleus. In this context, the total energy before the interaction (the sum of the gamma ray's energy and the nucleus's rest energy) equals the total energy after the interaction (the rest energy of the now excited nucleus). The specific equation used is: \[ E_{\text{p}} + mc^2 = E_{m^*} \] This equation ensures that the energy supplied by the gamma ray is accounted for in the energy of the excited nucleus.

Conservation of Momentum

Conservation of momentum is another fundamental principle in physics which states that the momentum of a closed system remains constant if no external forces act on it. When a gamma ray is absorbed by a nucleus, the system's total momentum before absorption must equal the total momentum after absorption. For a gamma ray (which has energy but no rest mass), its momentum is given by its energy divided by the speed of light: \[ \frac{E_{\text{p}}}{c} = p_{m^*} \] This momentum is then transferred to the nucleus, making the equation: \[ p_{m^*} = \frac{(E_{m^*}^2 - m^{*2}c^4)^{1/2}}{c} \]Here, the momentum of the excited nucleus is connected to its energy and rest mass.

Gamma Rays

Gamma rays are a type of electromagnetic radiation with very high energy. Unlike visible light, gamma rays have much higher frequencies and shorter wavelengths, which give them significant energy. When a gamma ray interacts with a nucleus, as in the provided exercise, it can be absorbed, causing the nucleus to move to an excited state. The involved energy transitions play a crucial role in nuclear reactions and can be precisely calculated using conservation laws.

Nuclear Mass and Energy

In nuclear physics, mass and energy are closely related through Einstein's famous equation \( E = mc^2 \). This relationship means that changes in the nuclear mass due to high-energy interactions (like absorbing a gamma ray) correspond to changes in energy. When a nucleus absorbs a gamma ray, its mass increases slightly to form an excited nucleus. This is represented as: \[ m^{*} = (2\frac{E_{\text{p}}}{c^2}m + m^2)^{1/2} \]This formula shows the dependence of the mass of the excited nucleus on the energy of the gamma ray and the original mass of the nucleus.

Limiting Cases in Physics

Exploring limiting cases helps in understanding the behavior of physical systems under extreme conditions. For the given exercise, considering very small gamma ray energies \(E_{\text{p}}\) compared to the nucleus's rest energy (\(mc^2\)) simplifies our analysis. When \(E_{\text{p}}\) is negligible, the excited nucleus’s mass \(m^{*}\) approaches the original mass \(m\). This makes sense because a low-energy gamma ray does not significantly alter the nucleus's mass, confirming that: \[ m^{*} = m \]Such analyses reveal the behavior of physical systems in different scenarios, reinforcing our understanding of fundamental principles.