Problem 31 - Physics for Scientists and Engineers with Modern Physics

Question

(II) The nuclide \(\frac{32}{15} \mathrm{P}\) 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) The daughter nucleus is \( \frac{32}{16} \mathrm{S} \). (b) The daughter's atomic mass is approximately 31.97207 u.

Step 1: Understand Beta Decay

The problem involves the decay of the nuclide \( \frac{32}{15} \mathrm{P} \). This is a beta decay process, where a neutron is converted into a proton, emitting a beta particle (electron) and an antineutrino. When a beta particle is emitted, the atomic number increases by 1, while the mass number remains unchanged.

Step 2: Identify the Daughter Nucleus

Since the original nuclide \( \frac{32}{15} \mathrm{P} \) decays by beta emission, increasing the atomic number by 1, the atomic number of the daughter nuclide will be 16 while the mass number remains 32. Therefore, the daughter nucleus is \( \frac{32}{16} \mathrm{S} \), which is sulfur.

Step 3: Determine the Daughter's Atomic Mass

The atomic masses of the elements around sulfur are typically close to their mass numbers when expressed in atomic mass units (u). However, to find the precise atomic mass, one would usually refer to a detailed atomic mass table. For sulfur-32, the typical atomic mass is approximately 31.97207 u.

Key Concepts

Nuclear Physics

Nuclear physics is a branch of physics that deals with the study of atomic nuclei and their interactions. It explores phenomena that occur in atomic nuclei, such as beta decay, fission, and fusion. In beta decay, a neutron in an atomic nucleus is transformed into a proton, accompanied by the emission of a beta particle (electron) and an almost undetectable neutrino, specifically an antineutrino. This process is fundamental in nuclear stability and transmutation of elements.
Beta decay affects the atomic number of the nucleus but not the mass number. The transformation results in a new element, known as the daughter nucleus, having one more proton than the parent. This is why understanding beta decay is crucial in nuclear physics.

• Beta particles are high-energy, high-speed electrons emitted from the nucleus.
• Antineutrinos are neutral and carry away some energy from the decay.
• The process is essential for determining the stability of isotopes.

Researchers in nuclear physics aim to understand the fundamental forces that hold the nucleus together, as well as the processes that cause changes within the nucleus, such as during beta decay.

Atomic Mass Unit

The atomic mass unit (u) is a standard unit used to express atomic and molecular weights, based on one twelfth of the mass of a carbon-12 atom. It provides a convenient way to describe the masses of atoms, which are incredibly small. An atomic mass unit is essential for understanding nuclear processes, like beta decay, as it helps quantify changes in mass.
In nuclear reactions, such as the decay of phosphorus-32 into sulfur-32, we express the mass of the daughter nucleus in atomic mass units. This allows for precision and consistency in calculations. Typically, the atomic mass of a nuclide closely aligns with its mass number because the mass is primarily concentrated in protons and neutrons.

• 1 atomic mass unit is approximately equal to 1.66053906660 × 10^-27 kilograms.
• It simplifies comparisons of the masses of different atoms and isotopes.
• Atomic mass tables provide precise values for detailed studies.

Understanding and using atomic mass units is vital for calculating changes in nuclear reactions and predicting the products of such reactions.

Daughter Nucleus

In nuclear physics, when a radioactive parent nucleus undergoes decay, it transforms into a more stable daughter nucleus. This transformation involves the parent nucleus emitting particles or radiation. In the case of beta decay, the daughter nucleus has one more proton than the parent nucleus, but retains the same mass number.
The daughter nucleus resulting from such a decay is often a different element with distinct chemical properties. In the example of phosphorus-32 undergoing beta decay, the resulting daughter nucleus is sulfur-32. This transition highlights how nuclear processes can change the identity of an element by altering its atomic number while leaving its mass number unchanged.

• The concept of a daughter nucleus is key to understanding nuclear transmutation.
• Daughter nuclei may undergo further decay to achieve stability.
• Identifying daughter nuclei helps in tracking nuclear decay series.

The transformation into a daughter nucleus through beta decay illustrates the dynamic and transformational nature of nuclear physics.