Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Exam: CSIR-NET DEC 2011 · Marks: 2 · Difficulty: Hard
The carbon-14 activity of an old wood sample is found to be $14.2$ disintegrations $\mathrm{min^{-1}\,g^{-1}}$. Calculate age of old wood sample, if for a fresh wood sample carbon-14 activity is $15.3$ disintegrations $\mathrm{min^{-1}\,g^{-1}}$ ($t_{1/2}$ carbon-14 is $5730$ years), is:
(a)5,000 years
(b)4,000 years
(c)877 years
(d)617 years
Answer
Answer (as printed): D
Explanation
$r_0=15.3\,\mathrm{dpm}$; $r=14.2\,\mathrm{dpm}$. Rate at any time $\propto$ Number of atoms. $\dfrac{r_0}{r}=\dfrac{N_0}{N}=\dfrac{15.3}{14.2}$. $t=\dfrac{2.303\times t_{1/2}}{0.693}\log\dfrac{N_0}{N}=\dfrac{2.303\times 5730}{0.693}\log\dfrac{15.3}{14.2}=617\ \text{years}$