Problem 31 - Physics for Scientists and Engineers with Modern Physics

Question

(II) The nuclide \({ }_{15}^{32} \mathrm{P}\) decays by emitting an electron whose maximum kinetic energy can be \(1.71 \mathrm{MeV}\). (a) What is the daughter nucleus? (b) Calculate the daughter's atomic mass (in u).

Step-by-Step Solution

Answer

(a) Daughter nucleus: \(_{16}^{32}\)S. (b) Daughter's atomic mass: ~31.9734 u.

Step 1: Understand the Decay Process

The nuclide \(_{15}^{32} \text{P} \) decays by beta-minus decay, where a neutron is converted to a proton, emitting an electron (beta particle) and an antineutrino. In beta-minus decay, the atomic number increases by 1 while the mass number remains constant.

Step 2: Identify the Daughter Nucleus

Since the nuclide \( _{15}^{32} \text{P} \) emits a beta particle, one of its neutrons is converted into a proton. This increases the atomic number by 1, resulting in: \[_{16}^{32} \text{S}\]. This indicates that the daughter nucleus is sulfur (S) with the same mass number.

Step 3: Calculate the Daughter's Atomic Mass

In beta decay, the change in nuclear mass is negligible, meaning the daughter's atomic mass is approximately the same as that of the parent minus the mass of the emitted electron. The mass of \( _{15}^{32} \text{P} \) is approximately 31.9739 u, and the mass of an electron is about 0.00054858 u. Thus, the atomic mass of the daughter \( _{16}^{32} \text{S} \) can be estimated as: \[ 31.9739 \, \text{u} - 0.00054858 \, \text{u} = 31.97335142 \, \text{u} \].

Key Concepts

Nuclear Reactions

Nuclear reactions involve the change in an atom's nucleus, leading to the transformation of one element into another. These reactions are different from chemical reactions, where only electrons outside the nucleus participate. In nuclear reactions, particles such as protons, neutrons, and electrons, known as beta particles, play a key role.
In this context, beta decay is a common type of nuclear reaction. During beta-minus decay, a neutron inside the nucleus is transformed into a proton. This transformation occurs because the neutron emits an electron and an antineutrino. As a result, the atomic number increases by one, converting the element into another element, while the mass number remains unchanged.
To identify the new element, or daughter nucleus, focus on the increase of the atomic number by one. This minor change makes all the difference; for example, phosphorus with an atomic number of 15 becomes sulfur with an atomic number of 16.

Atomic Mass Calculation

Atomic mass is crucial in understanding both the characteristics of an element and how it behaves in nuclear reactions. When a nuclide undergoes beta decay, its atomic mass barely changes, because the mass of the emitted electron is extremely small.
The atomic mass unit (u) is used to express tiny changes in mass. For instance, the mass of the phosphorus nuclide \(_{15}^{32} \text{P} \) is approximately 31.9739 u. During beta decay, a small mass, about 0.00054858 u, is lost due to the emitted electron. Therefore, the atomic mass of the daughter nuclide, sulfur \(_{16}^{32} \text{S} \), is calculated as follows:

• Start with the original mass of phosphorus: 31.9739 u
• Subtract the mass of the electron: 31.9739 u - 0.00054858 u = 31.97335142 u

This calculation shows how little the atomic mass changes, supporting the notion that atomic mass is fairly conserved in beta decay.

Nuclide Transformation

Nuclide transformation is the process by which one nuclide changes into another during nuclear reactions. This transformation is pivotal in understanding radioactive decay and the behavior of elements under different conditions.
In beta-minus decay, the parent nuclide undergoes transformation due to the conversion of a neutron into a proton. This results in a change in the element itself, from phosphorus \(_{15}^{32} \text{P} \) to sulfur \(_{16}^{32} \text{S} \). Importantly, the mass number remains constant at 32, as the change involves only subatomic particles, not the overall mass of the nucleus.
Understanding these transformations helps in predicting the formation of daughter nuclei and assessing the types of particles that are emitted. This insight is useful in various applications, such as nuclear medicine and energy production, where control of nuclear reactions and knowledge of resulting isotopes are vital.