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Chapter 5: Problem 90

The radioisotope sodium- 24 is used to determine the levels of electrolytes in the body. A \(16-\mu \mathrm{g}\) sample of sodium- 24 decays to \(2.0 \mu \mathrm{g}\) in \(45 \mathrm{~h}\). What is the half-life, in hours, of sodium- \(24 ?(5.4)\)

Short Answer

Expert verified
The half-life of sodium-24 is 15 hours.

Step by step solution

01

- Understand Decay Formula

The decay of a radioactive isotope is given by the formula: \[ N(t) = N_0 \times \frac{1}{2}^{t / t_{1/2}} \] where \( N(t) \) is the remaining quantity at time \( t \), \( N_0 \) is the initial quantity, and \( t_{1/2} \) is the half-life.
02

- Identify Given Values

Here, the initial quantity \( N_0 \) is \( 16 \, \text{渭g} \), the remaining quantity \( N(t) \) is \( 2.0 \, \text{渭g} \), and the time \( t \) is \( 45 \, \text{hours} \). We need to find the half-life \( t_{1/2} \).
03

- Set Up Equation with Given Values

Using the formula: \[ 2.0 = 16 \times \frac{1}{2}^{45 / t_{1/2}} \] Divide both sides by 16 to isolate the exponential term: \[ \frac{2.0}{16} = \frac{1}{2}^{45 / t_{1/2}} \] Simplify: \[ \frac{1}{8} = \frac{1}{2}^{45 / t_{1/2}} \]
04

- Solve for Half-Life Exponent

Rewrite \( \frac{1}{8} \) as \( \frac{1}{2^3} \): \[ \frac{1}{2^3} = \frac{1}{2}^{45 / t_{1/2}} \] Therefore, \( 3 = \frac{45}{t_{1/2}} \).
05

- Solve for Half-Life

Rearrange to solve for \( t_{1/2} \): \[ t_{1/2} = \frac{45}{3} = 15 \text{ hours} \]

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Key Concepts

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

radioactive decay
Radioactive decay is a natural process where an unstable atomic nucleus loses energy by emitting radiation. This can include the emission of particles like alpha particles, beta particles, or gamma rays. Over time, this process reduces the number of unstable nuclei in the sample, and the substance gradually transforms into a more stable form. Each type of radioactive isotope decays at its own unique rate, which is often expressed in terms of its half-life.
decay formula
The decay formula is used to calculate the remaining quantity of a radioactive substance after a certain period of time. The general formula for radioactive decay is given by:
\[ N(t) = N_0 \times \frac{1}{2}^{t / t_{1/2}} \] where:
  • \
sodium-24
Sodium-24 is a radioactive isotope of sodium commonly used in the field of medicine and other scientific applications.
For instance, it helps in determining electrolyte levels in the human body, making it valuable in diagnostic tests. Sodium-24 decays by emitting beta particles, which can be tracked to gather data about various physiological processes. With a half-life calculated in the above exercise to be approximately 15 hours, sodium-24 decays relatively quickly. This makes it suitable for short-term medical applications where prolonged radioactivity could be harmful.

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Most popular questions from this chapter

a. The dosage of technetium-99m for a lung scan is 20\. \(\mu \mathrm{Ci} / \mathrm{kg}\) of body mass. How many millicuries of technetium-99m should be given to a \(50.0-\mathrm{kg}\) person $$ (1 \mathrm{mCi}=1000 \mu \mathrm{Ci}) ? $$ b. A person receives 50 rad of gamma radiation. What is that amount in grays?

Write the balanced nuclear equation for each of the following: (5.1,5.2) a. \(\mathrm{Th}-225(\alpha\) decay \()\) b. \(\mathrm{Bi}-210(\alpha\) decay \()\) c. cesium-137 ( \(\beta\) decay) d. \(\operatorname{tin}-126(\beta\) decay \()\) e. \(\mathrm{F}-18\left(\beta^{+}\right.\) emission \()\)

A 120 -mg sample of technetium- \(99 m\) is used for a diagnostic test. If technetium-99m has a half-life of \(6.0 \mathrm{~h}\), how many milligrams of the technetium-99m sample remains active \(24 \mathrm{~h}\) after the test? (5.4)

Indicate whether each of the following is characteristic of the fission or fusion process, or both: a. Very high temperatures are required to initiate the reaction. b. Less radioactive waste is produced. c. Hydrogen nuclei are the reactants. d. Large amounts of energy are released when the nuclear reaction occurs.

A nuclear technician was accidentally exposed to potassium- 42 while doing brain scans for possible tumors. The error was not discovered until 36 h later when the activity of the potassium- 42 sample was \(2.0 \mu \mathrm{Ci}\). If potassium- 42 has a half-life of \(12 \mathrm{~h}\), what was the activity of the sample at the time the technician was exposed? (5.3,5.4)
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