Question

Half-life for radioactive ${}^{14}C$ is 5760 yr. In how many years, 200 mg of ${}^{14}C$ will be reduced to 25 mg?

• A. 5760 yr
• B. 11520 yr
• C. 17280 yr
• D. 23040 yr

Answer 1

Answer: (C)

Half-life of radioactive
$t_{1/2}25760yr,[R{]}_0=200mg$
[R] = 25 mg
Rate constant (k) =$\frac{0.6932}{t_{1/2}}=\frac{0.6932}{5760}y{r}^{-1}$
$\therefore k=\frac{2.303}{t}log\frac{|R{|}_0}{|R|}$
$(R_0$ = 200 mg inital amount)
$\frac{0.6932}{t}=\frac{2.303}{t}log\frac{200}{25}$
(R = 25 mg reduced amount)
$=\frac{2.303}{t}$ log 8
$\frac{0.6932}{5760}=\frac{2.303}{t}\times 0.9030$
$t=\frac{5760\times 2.303\times 0.9030}{0.6932}$
$=\frac{11978.54}{0.6932}=17280yr$