Problem 75 - Physics for Scientists and Engineers with Modern Physics

Question

The nuclide \(\frac{191}{76} \mathrm{Os}\) decays with \(\beta^{-}\) energy of 0.14 MeV accompanied by \(\gamma\) rays of energy 0.042 MeV and 0.129 MeV. \((a)\) What is the daughter nucleus? \((b)\) Draw an energy-level diagram showing the ground states of the parent and daughter and excited states of the daughter. (c) To which of the daughter states does \(\beta^{-}\) decay of \(\frac{191}{76} \mathrm{Os}\) occur?

Step-by-Step Solution

Answer

(a) The daughter nucleus is \(^191_{77}\text{Ir}\). (b)/(c) Draw the energy levels, with \(\beta^{-}\) decay to Ir's ground state.

Step 1: Identify the Decay Process

The problem states that we have a beta decay. In a beta-minus (\(\beta^{-}\)) decay, a neutron in the nucleus is transformed into a proton, and an electron (the beta particle) and an antineutrino are emitted. Thus, the atomic (proton) number increases by 1, while the mass number remains unchanged.

Step 2: Determine the Daughter Nuclide

The parent nuclide is given as \( ^{191}_{76} \text{Os} \). Applying the beta-minus decay process, the atomic number increases from 76 to 77 while the mass number stays 191. Hence, the daughter nuclide will be \( ^{191}_{77} \text{Ir} \), iridium.

Step 3: Draw the Energy-Level Diagram

1. Start by illustrating the ground state of the parent nuclide \( ^{191}_{76} \text{Os} \).2. Indicate the ground state of the daughter nuclide \( ^{191}_{77} \text{Ir} \) below the level of Os.3. Display excited states of Ir as energy levels above its ground state, using the provided \( \gamma \)-ray energies. The initial excited state is at 0.042 MeV and the higher excited state is at 0.129 MeV above the ground state.

Step 4: Determine the Beta Decay Transition

The emission of a \( \beta^{-} \) particle with 0.14 MeV energy suggests a transition of \( \text{Os} \) either directly to the ground state of \( \text{Ir} \) or potentially to an excited state. Calculating the energy, if the \( \beta^{-} \) particle is emitted to reach directly a specific state, the energy must match either the ground state or an excited state (taking into account the gamma transitions). Based on the given decay scheme, \( \beta^{-} \) decay occurs primarily to the ground state of \( ^{191}_{77} \text{Ir} \).

Key Concepts

beta-minus decay

Beta-minus decay is a type of nuclear decay where a neutron within the nucleus transforms into a proton. During this process, an electron (also known as a beta particle) and an antineutrino are emitted from the nucleus. One of the key outcomes of beta-minus decay is that the atomic number, which represents the number of protons in an atom, increases by one. However, the mass number, which is the total count of protons and neutrons, remains unchanged. For instance, as seen in the decay of \[^{191}_{76} \text{Os}\] (osmium), a neutron turns into a proton, resulting in the formation of \[^{191}_{77} \text{Ir}\] (iridium). This transformation increases the atomic number from 76 to 77, showing the essence of beta-minus decay.

daughter nuclide

When a parent nuclide undergoes nuclear decay, it turns into a different element, often referred to as the daughter nuclide. In beta-minus decay, the daughter nuclide will have an atomic number that is one unit higher than that of the parent nuclide, since a neutron converts into a proton. In the example of \[^{191}_{76} \text{Os}\], this decay results in the daughter nuclide \[^{191}_{77} \text{Ir}\]. The mass number remains consistent in both the parent and daughter nuclides, so they both have a mass number of 191. The concept of a daughter nuclide is crucial, as it manifests the physical transformation of matter during nuclear decay processes.

energy-level diagram

An energy-level diagram illustrates the various energy states of the particles involved in a nuclear process. In the case of the decay of \[^{191}_{76} \text{Os}\], such a diagram helps visualize how energy levels change as beta-minus decay occurs.

• Begin with the ground state of the parent nuclide, osmium (Os), at a certain energy level.


• The ground state of the daughter nuclide, iridium (Ir), needs to be depicted, as it reflects the stable state after decay.


• Above Ir's ground state, excited states are depicted, which arise due to gamma-ray emissions.

In this scenario, the \(\gamma\)-rays have energies of 0.042 MeV and 0.129 MeV. These values help position the excited states of iridium in the diagram. By illustrating both ground and excited states of Ir, the energy-level diagram provides a clear visual representation of the transitions occurring during the nuclear decay process.

gamma-ray emission

Gamma-ray emission is a form of electromagnetic radiation often accompanying nuclear decay such as beta-minus decay. After the initial decay, the daughter nucleus may be in an excited state. It releases gamma rays to move from these higher energy (excited) states to a lower energy state or ground level. These emissions are photons with very high energy and can be detected even when the particles themselves are not.

• In this example, the emission of gamma rays with energies 0.042 MeV and 0.129 MeV signals transitions between specific energy levels of the daughter nucleus \( ^{191}_{77} \text{Ir}\).


• These transitions often occur right after beta-minus decay has occurred, as the daughter nuclide sheds excess energy.


• The gamma-ray energy values tell us which excited states are involved and these are important when constructing energy-level diagrams to show how energy states change.

Understanding gamma-ray emission helps interpret and predict nuclear transformation events, making it an essential concept in nuclear physics.