Question

Calculate the weight in grams of 1 Ci of ${}^{14}C(T_{50}=5000yr)$,$(1yr=31.536\times {10}^{6}s)$.
Plan $-\frac{dN}{dt}=\lambda N=\lambda \frac{w}{A}{N}_0$,where w is the weight of the element of mass number A; $N_0$ is the Avogadro's number.

• A. $w=0.196$ g
• B. $w=0.796$ g
• C. $w=0.896$ g
• D. None of these

Answer 1

The correct option is A $w=0.196$ g

Let weight required $=wg$
$=\frac{w}{14}\times 6.02\times {10}^{23}$ atoms
Disintegration constant $\left(\lambda \right)=\frac{0.693}{5000\times 31.536\times {10}^6}{s}^{-1}$
and $1Ci=3.7\times {10}^{10}$ dps
$-\frac{dN}{dt}=\lambda N$
$3.7\times {10}^{10}=\frac{0.693\times w\times 6.02\times {10}^{23}}{5000\times 31.536\times {10}^6\times 14}$
$w=0.196$ g