Problem 46 - Fundamentals of Physics

Question

The radioactive nuclide \({ }^{99}\) Tc can be injected into a patient's bloodstream in order to monitor the blood flow, measure the blood volume, or find a tumor, among other goals. The nuclide is produced in a hospital by a "cow" containing \({ }^{99} \mathrm{Mo},\) a radioactive nuclide that decays to \({ }^{99} \mathrm{Tc}\) with a half-life of \(67 \mathrm{~h}\). Once a day, the cow is "milked" for its \({ }^{99} \mathrm{Tc},\) which is produced in an excited state by the \({ }^{99} \mathrm{Mo} ;\) the \({ }^{99} \mathrm{Tc}\) de- excites to its lowest energy state by emitting a gamma-ray photon, which is recorded by detectors placed around the patient. The de-excitation has a half- life of \(6.0 \mathrm{~h}\). (a) By what process does \({ }^{99}\) Mo decay to \({ }^{99} \mathrm{Tc} ?\) (b) If a patient is injected with an \(8.2 \times 10^{7}\) Bq sample of \({ }^{99} \mathrm{Tc}\), how many gamma-ray photons are initially produced within the patient each second? (c) If the emission rate of gamma-ray photons from a small tumor that has collected \({ }^{99} \mathrm{Tc}\) is 38 per second at a certain time, how many excited-state \({ }^{99} \mathrm{Tc}\) are located in the tumor at that time?

Step-by-Step Solution

Answer

(a) \(^{99}\) Mo decays by beta decay. (b) 8.2 x 10^7 gamma-rays/s. (c) About 1.18 x 10^6 excited \(^{99} \mathrm{Tc}\) nuclei.

Step 1: Determine the Decay Process

To identify the process by which \(^{99}\) Mo decays to \(^{99}\) Tc, we recognize the common decay methods. \(^{99}\) Mo decays by beta decay, where a neutron is transformed into a proton, an electron (beta particle), and an antineutrino.

Step 2: Understand Initial Sample Activity

The activity (measured in Bq) reflects the number of decays per second. Given \(8.2 \times 10^{7}\) Bq for the \(^{99} \mathrm{Tc}\) sample, the initial emission of gamma-ray photons equals this activity. Thus, \(8.2 \times 10^{7}\) gamma-rays are produced per second.

Step 3: Relate Decay Events to Excited-State Nuclei

Each decay event of excited-state \(^{99} \mathrm{Tc}\) leads to the emission of a gamma-ray photon. The rate of 38 gamma emissions per second indicates 38 excited \(^{99} \mathrm{Tc}\) nuclei decay to the ground state per second.

Step 4: Calculate Number of Excited-State Tc Nuclei

Using the half-life (\(T_{1/2} = 6.0\) hours) of the excited-state \(^{99} \mathrm{Tc}\), we find the decay constant using \(\lambda = \frac{0.693}{T_{1/2}}\). Substitute \(T_{1/2}\,\) in seconds: \(\lambda = \frac{0.693}{21600}\,\) s.Number of excited nuclei \(= \frac{38}{\lambda}\) \(\approx \frac{38}{3.21 \times 10^{-5}} \approx 1.18 \times 10^{6}\,e\) excited \(^{99} \mathrm{Tc}\) nuclei in the tumor.

Key Concepts

Beta Decay

In the world of radioactive decay, beta decay is a fascinating process. It involves the transformation of a neutron into a proton within the nucleus of an atom. As a result, an electron, known as a beta particle, is ejected from the nucleus. Alongside the electron, an antineutrino, which is a very light and almost non-interacting particle, is also produced.
Beta decay is pivotal because it changes the identity of the atom. In the case of the radioactive nuclide \(^{99} \text{Mo} \), this process transforms it into \(^{99} \text{Tc} \). This change is crucial for medical applications, where \(^{99} \text{Tc} \) is used to trace blood flow and detect anomalies. It's an essential decay process that not only alters the atom itself but has a profound impact in the world of medicine.

Gamma-ray Emission

Gamma-ray emission is a phenomenon where an excited atomic nucleus releases excess energy. This energy is emitted in the form of gamma rays, which are high-energy photons. Unlike other types of decay, gamma-ray emission does not change the number of protons or neutrons in the nucleus.
This process is important for applications such as medical imaging. When \(^{99} \text{Tc} \) decays back to its ground state, it emits gamma-ray photons. These photons are detected by instruments surrounding the patient, allowing doctors to see images of what's happening inside the body. Gamma rays have the advantage of penetrating tissues with ease, making them highly effective for scanning and diagnostic purposes.

Half-life

The term half-life describes the time it takes for half of a sample of a radioactive substance to decay. Every radioactive nuclide has a unique half-life, characterizing its rate of decay.
For example, in the given exercise, \(^{99} \text{Mo} \) transforms into \(^{99} \text{Tc} \) with a half-life of 67 hours. This means that in 67 hours, half of the initial quantity of \(^{99} \text{Mo} \) will have decayed into \(^{99} \text{Tc} \). Furthermore, the excited \(^{99} \text{Tc} \) itself has a separate half-life of 6.0 hours to decay and emit gamma-ray photons.
Half-lives are fundamental for nuclear physics and practical computations, for instance, determining how long a sample remains sufficiently active for medical procedures.

Radioactive Nuclide

A radioactive nuclide, sometimes called a radioisotope, is an atom with an unstable nucleus. Such instability leads to the atom releasing energy in the form of radiation.
In the discussed problem, both \(^{99} \text{Mo} \) and \(^{99} \text{Tc} \) are examples of radioactive nuclides. \(^{99} \text{Mo} \) is the precursor that decays into \(^{99} \text{Tc} \), a nuclide used extensively in medical imaging to detect abnormal regional blood flow or tumors.
Understanding radioactive nuclides is crucial in a variety of fields from medicine to power generation, as they help understand both the behavior of elements under radioactive decay and the management of their applications safely and effectively.