A wooden artifact and a freshly cut tree have | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The decay of carbon follows first-order kinetics. Given: Initial activity $(r_o)$ = $15.2 \ min^{-1} g^{-1}$,Final activity $(r)$ = $7.6 \ min^{-1

Vedclass · Accepted Answer

$5760$ years

Vedclass · Answer

(A) The decay of carbon follows first-order kinetics. Given: Initial activity $(r_o)$ = $15.2 \ min^{-1} g^{-1}$,Final activity $(r)$ = $7.6 \ min^{-1} g^{-1}$,and $t_{1/2} = 5760$ years. Since $r = \frac{r_o}{2}$,the artifact has undergone exactly one half-life. Therefore,the age of the artifact $(t)$ = $1 	imes t_{1/2} = 5760$ years. Alternatively,using the formula $t = \frac{t_{1/2}}{0.693} \ln(\frac{r_o}{r}) = \frac{5760}{0.693} \ln(\frac{15.2}{7.6}) = \frac{5760}{0.693} \ln(2) = 5760$ years.

$A$ wooden artifact and a freshly cut tree have carbon activity of $7.6$ and $15.2 \ min^{-1} g^{-1}$ respectively. Given the half-life $(t_{1/2})$ of carbon is $5760$ years,the age of the artifact is:

Similar Questions

The half-life of $_{6}^{14}C$ having a decay constant of $2.25 \times 10^{-4} \text{ year}^{-1}$ is .......... years.

The radioactivity due to the $^{14}C$ isotope (half-life $6000 \ years$) of a sample of wood from an ancient tomb was found to be nearly half that of fresh wood. The tomb is therefore about ........... years old.

The radioactivity of a radioactive element becomes $\frac{1}{10}$ of the original radioactivity after $2.303 \ s$. The half-life period is: (in $s$)

The amount of $_{53}I^{128}$ $(t_{1/2} = 25 \ min)$ remaining after $75 \ min$ is:

The half-life period of a radioactive metal is $20 \ days$. What fraction of the metal remains after $80 \ days$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module