Question

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 of sodium- 24 ?

Short answer

The half-life of sodium-24 is approximately 15 hours.

Step-by-step solution

- Understand the Decay Formula

The decay of a substance over time can be modeled using the formula: \[ N(t) = N_0 \times \frac{1}{2}^{\frac{t}{T_{1/2}}}\] where \( N(t) \) is the amount remaining after time \( t \), \( N_0 \) is the initial amount, and \( T_{1/2} \) is the half-life.

- Identify Given Values

Identify the values given in the problem: Initial amount \( N_0 = 16 \, \text{μg} \) Remaining amount \( N(t) = 2.0 \, \text{μg} \) Time \( t = 45 \, \text{h} \).

- Set Up the Decay Equation

Substitute the known values into the decay formula: \[ 2.0 = 16 \times \frac{1}{2}^{\frac{45}{T_{1/2}}} \]

- Simplify the Equation

Divide both sides by 16: \[ \frac{2.0}{16} = \frac{1}{2}^{\frac{45}{T_{1/2}}} \] This simplifies to: \[ 0.125 = \frac{1}{2}^{\frac{45}{T_{1/2}}} \]

- Take the Logarithm

Take the natural logarithm (ln) of both sides to solve for the exponent: \[ \text{ln}(0.125) = \text{ln}\bigg(\frac{1}{2}^{\frac{45}{T_{1/2}}}\bigg) \] Using the property of logarithms: \[ \text{ln}(0.125) = \frac{45}{T_{1/2}} \text{ln}\bigg(\frac{1}{2}\bigg) \]

- Solve for Half-Life \( T_{1/2} \)

Calculate the natural logarithms: \[ \text{ln}(0.125) = -2.0794 \] \[ \text{ln}\bigg(\frac{1}{2}\bigg) = -0.6931 \] Substitute these back into the equation: \[ -2.0794 = \frac{45}{T_{1/2}} \times (-0.6931) \] Simplify to get: \[ T_{1/2} = \frac{45 \times 0.6931}{2.0794} \]

- Final Calculation

Complete the final calculation: \[ T_{1/2} \approx 15 \text{ hours} \]

Key concepts

Radioactive Decay

Radioactive decay is a process where unstable atomic nuclei release energy by emitting radiation. This continues until they transform into more stable products. Different types of radiation include alpha, beta, and gamma rays. This transformation is natural and spontaneous. Over time, the number of radioactive atoms in a sample decreases. Key points to understand:

• The rate of decay is proportional to the number of undecayed atoms present.
• The half-life is the period it takes for half of the sample to decay.
• The decay can be represented mathematically, making calculations predictable and easier.

This concept is essential for understanding how substances like sodium-24 change over time.

Natural Logarithm

Natural logarithms simplify the computation of decay problems. They’re logarithms with the base e, where e is an irrational constant approximately equal to 2.71828. The natural logarithm is denoted as ln. In decay problems, natural logarithms help us solve exponential equations. Here’s how:

• By taking the natural logarithm of both sides of an equation, we can simplify and solve for unknown variables.
• Properties such as ln(a^b) = b*ln(a) are particularly useful.

For our problem, we start with the decay formula. Taking the natural logarithm helps isolate and find the half-life. It’s a potent tool for transforming multiplicative relationships into additive ones.

Sodium-24 Decay

Sodium-24 is a radioactive isotope with a known half-life. It's commonly used in medical and industrial applications. The decay of Sodium-24 involves it transforming into another element, releasing energy in the process. In our exercise:

• We start with an initial amount of 16 μg of sodium-24.
• We observe that it decays to 2.0 μg over 45 hours.

By using the decay formula and properties of natural logarithms, we can determine the half-life of sodium-24. This process illustrates the practical application of the concepts. Calculations like these help us safely handle and use radioactive materials in various industries.