Question

A nuclear explosion has taken place leading to increase in the concentration of $C^{14}$ in nearby areas. $C^{14}$ concentration is $C_1$ in nearby areas and $C_2$ in areas far away. If the age of the fossil is determined to be $T_1$ and $T_2$ at the places respectively then :

• A. The age of the fossil will increase at the place where explosion has taken place and $T_1-T_2=\frac{1}{\lambda }\ln \frac{C_1}{C_2}$.
• B. The age of the fossil will decrease at the place where explosion has taken place and $T_1-T_2=\frac{1}{\lambda }\ln \frac{C_1}{C_2}$.
• C. The age of fossil will be determined to be same.
• D. $\frac{T_1}{T_2}=\frac{C_1}{C_2}$

Answer 1

Answer: (A)

The age of the fossil will increase at the place where explosion has taken place and $T_1-T_2=\frac{1}{\lambda }\ln \frac{C_1}{C_2}$.

We can mathematically quantify the rate of radioactive decay.

$N=N_0{e}^{-\lambda t}$

$N_0$ represents the original number of nuclei at time t=0

$N$ represents no of nuclei at time t

Concentration is $\frac{N_0}{N}=C$

Given at time $T_1$, Concentration will be $C_1$

Given at time $T_2$, Concentration will be $C_2$

From the above formula, we get:

$\frac{C_1}{C_2}=\frac{e^{\lambda T_1}}{e^{\lambda T_2}}$

$⟹\ln \frac{C_1}{C_2}=\lambda (T_1-T_2)$

$⟹\frac{1}{\lambda }\ln \frac{C_1}{C_2}=T_1-T_2$

The age of the fossil will increase with an increase in concentration from the above equation.