Question

If a radioactive substance decays 10% in every 16 hours, then the percentage of the radioactive substance that remains after 2 days is

• A. 82.2
• B. 18.8
• C. 27.1
• D. 72.9

Hint:

For radioactive decay problems involving fractions or percentages over time, you often don't need to calculate the decay constant $\lambda$ explicitly. Use the property $(e^{-\lambda t_1})^{t_2/t_1} = e^{-\lambda t_2}$. If a fraction $f$ remains after time $t_1$, then after time $n \cdot t_1$, the fraction remaining will be $f^n$.

Answer 1

The Correct Option is D

Let $N_0$ be the initial amount of the radioactive substance.
The law of radioactive decay is given by $N(t) = N_0 e^{-\lambda t}$, where $\lambda$ is the decay constant.
We are given that the substance decays by 10% in 16 hours. This means 90% remains.
Let $t_1 = 16$ hours. Then $N(t_1) = 0.90 N_0$.
$0.90 N_0 = N_0 e^{-\lambda(16)}$.
$0.9 = e^{-16\lambda}$.
We need to find the percentage that remains after 2 days.
Time $t_2 = 2$ days = $2 \times 24$ hours = 48 hours.
We need to find $\frac{N(t_2)}{N_0} = \frac{N_0 e^{-\lambda(48)}}{N_0} = e^{-48\lambda}$.
We can rewrite this using the information we have:
$e^{-48\lambda} = e^{-16\lambda \times 3} = (e^{-16\lambda})^3$.
From the first part, we know $e^{-16\lambda} = 0.9$.
So, the fraction remaining after 48 hours is $(0.9)^3$.
$(0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.9 = 0.729$.
The percentage remaining is $0.729 \times 100% = 72.9%$.