What is the half-life of a 12 g sample of radioisotope that decayed to 6 g in 28 years?A.7 yearsB.28 yearsC.56 yearsD.14 years

1. Break Down the Problem

To find the half-life of the radioisotope, we will analyze the decay of the sample over the given time period. The problem states that a 12 g sample decayed to 6 g in 28 years.

2. Relevant Concepts

The half-life ($t_{1/2}$) is the time taken for half of the radioactive sample to decay. The decay can be modeled using the formula: $N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}}$ where:

• $N(t)$ = remaining amount of substance after time $t$
• $N_0$ = initial amount of substance
• $t$ = time elapsed
• $t_{1/2}$ = half-life

3. Analysis and Detail

From the problem:

• Initial amount ($N_0$) = 12 g
• Remaining amount ($N(t)$) = 6 g
• Time elapsed ($t$) = 28 years

Substituting the values into the equation: $6 = 12 \left( \frac{1}{2} \right)^{\frac{28}{t_{1/2}}}$

Dividing both sides by 12: $\frac{1}{2} = \left( \frac{1}{2} \right)^{\frac{28}{t_{1/2}}}$

Since the bases are the same, we set the exponents equal: $1 = \frac{28}{t_{1/2}}$ Solving for $t_{1/2}$: $t_{1/2} = 28 \text{ years}$

4. Verify and Summarize

The calculations confirm that the half-life of the radioisotope is indeed 28 years since the sample reduced from 12 g to 6 g in that time frame.

Final Answer

The half-life of the radioisotope is 28 years (B).