Question

20 mg of C$^{14}$ has a half-life of 5760 years, 100 mg of sample containing C$^{14}$, is reduced to 25 mg in :

• A. 57600 years
• B. 1440 years
• C. 17280 years
• D. 11520 years

Answer 1

Answer: (D)

11520 years

Answer 2

For radioactive decay,

$N = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}$

With $N_0=100$ mg, $N=25$ mg, and $T_{1/2}=5760$ years,

$25 = 100 \left(\frac{1}{2}\right)^{t/5760} \quad \Rightarrow \quad \left(\frac{1}{2}\right)^{t/5760} = \frac{1}{4} = \left(\frac{1}{2}\right)^2$

Thus,

$\frac{t}{5760} = 2 \quad \Rightarrow \quad t=11520 \text{ years.}$