Question

The half-life period of $C^{14}$ is 5670 years. If $56g$ of $C^{14}$ was present initially, how many atoms of $C^{14}$ is left after 22680 years?

• A. $1.5\times {10}^{23}$
• B. $2.9\times {10}^{27}$
• C. $1.5\times {10}^{27}$
• D. $2.9\times {10}^{23}$

Answer 1

The correct option is A $1.5\times {10}^{23}$

$t_{1/2}=5670\lambda =\frac{1}{t_{1/2}}k=\lambda =\frac{1}{5670}\ln \frac{A}{A_o}=-kt{A}_o\to \frac{50}{14}\times 6.02\times {10}^{23}{A}_o=4\times 6.022\times {10}^{23}\ln \frac{A}{A_o}=-\frac{1}{5670}\times 22680\left(1\right)$

Solving equation (1) we get

$A=1.5\times {10}^{23}atoms$