Question

2 kg of a pure radioactive substance after a passage of \(N\) years is left with only 125 g of pure substance. The half-life of the radioactive substance is 12.5 y. Find the value of \(N\).

• A. 50 y
• B. 37.5 y
• C. 25 y
• D. 100 y

Hint:

Use \(N_t = N_0 (1/2)^{t/T_{1/2}}\) for decay calculations. Express remaining fraction as a power of 2 to solve for elapsed time.

Answer 1

The Correct Option is A

Step 1: Radioactive decay formula.
\[ N_t = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}} \] where \(N_0 = 2~\text{kg}\), \(N_t = 0.125~\text{kg}\), \(T_{1/2} = 12.5~\text{y}\).

Step 2: Substitute values.
\[ 0.125 = 2 \left(\frac{1}{2}\right)^{N/12.5} \]

Step 3: Simplify.
\[ \frac{0.125}{2} = \left(\frac{1}{2}\right)^{N/12.5} \Rightarrow \frac{1}{16} = \left(\frac{1}{2}\right)^{N/12.5} \]

Step 4: Express as power of 2.
\[ \frac{1}{16} = \frac{1}{2^4} \Rightarrow \left(\frac{1}{2}\right)^4 = \left(\frac{1}{2}\right)^{N/12.5} \]

Step 5: Solve for \(N\).
\[ N/12.5 = 4 \Rightarrow N = 50~\text{y} \]

Step 6: Conclusion.
The substance takes 50 years to decay from 2 kg to 125 g.