Question

A radioactive element of mass 1 kg after N years is left with only 125 g. If the half life of the element is 12.5 y, then the value of N is

• A. 37.5 years.
• B. 25.0 years.
• C. 50.0 years.
• D. 75.0 years.

Hint:

Radioactive decay law: $N_t = N_0 (1/2)^{t/T_{1/2}}$, where $N_t$ is quantity at time $t$, $N_0$ is initial quantity, $T_{1/2}$ is half-life. This applies to mass as well.
Determine the number of half-lives passed: $M_t/M_0 = (1/2)^n$, where $n = t/T_{1/2}$ is the number of half-lives.
In this case, $125\text{g}/1000\text{g} = 1/8 = (1/2)^3$. So, 3 half-lives have passed.
Total time $t = n \times T_{1/2}$.

Answer 1

The Correct Option is A

The formula for radioactive decay relating initial quantity ($M_0$), final quantity ($M_t$), half-life ($T_{1/2}$), and time elapsed ($t=N$ years) is: $M_t = M_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$. Given values: Initial mass $M_0 = 1 \text{ kg} = 1000 \text{ g}$. Final mass $M_t = 125 \text{ g}$. Half-life $T_{1/2} = 12.5 \text{ years}$. Time elapsed $t = N$ years. We need to find $N$. Substitute the values into the decay formula: $125 \text{ g} = 1000 \text{ g} \left(\frac{1}{2}\right)^{\frac{N}{12.5}}$. Divide by 1000 g: $\frac{125}{1000} = \left(\frac{1}{2}\right)^{\frac{N}{12.5}}$. Simplify the fraction $\frac{125}{1000}$: $\frac{125}{1000} = \frac{1 \times 125}{8 \times 125} = \frac{1}{8}$. So, $\frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{N}{12.5}}$. We know that $\frac{1}{8} = \left(\frac{1}{2}\right)^3$. Therefore, $\left(\frac{1}{2}\right)^3 = \left(\frac{1}{2}\right)^{\frac{N}{12.5}}$. Equating the exponents: $3 = \frac{N}{12.5}$. Now, solve for $N$: $N = 3 \times 12.5 \text{ years}$. $N = 37.5 \text{ years}$. This means that after $37.5$ years, the element is left with 125 g. This matches option (a). \[ \boxed{37.5 \text{ years}} \]