A nuclear explosion has taken place leading to an increase in the concentration of ${\mathrm{C}}^{14}$ in nearby areas. ${\mathrm{C}}^{14}$ concentration is ${\mathrm{C}}_1$ in nearby areas and ${\mathrm{C}}_2$ in areas far away. If the age of the fossil is determined to be ${\mathrm{T}}_1$ and ${\mathrm{T}}_2$ at the places respectively then:

$\left[\mathrm{Hint}:\mathrm{\lambda }=\frac{1}{{\mathrm{T}}_1}\ln \frac{{\mathrm{C}}_1}{\mathrm{X}}=\frac{1}{{\mathrm{T}}_2}\ln \frac{{\mathrm{C}}_2}{\mathrm{X}}\right]$

The edge length of a unit cell of a metal having molecular weight $75 \mathrm{g}/\mathrm{mol}$ is $5 {\mathrm{A}}^{\mathrm{o}}$ which crystallises in a cubic lattice. If the density is $2 \mathrm{g}/\mathrm{c}.\mathrm{c}.$, the radius of the metal atom in $\mathrm{pm}$ is:

[Hint: First calculate $\mathrm{Z}(=2)$ which means that the structure is b.c.c., for which $\mathrm{r}=\sqrt{3}/4·\mathrm{a}$ ; $1\stackrel{\mathrm{o}}{\mathrm{A}}={10}^{-10} \mathrm{m}$ and $1 \mathrm{m} = {10}^{12} \mathrm{pm}; 1\stackrel{\mathrm{o}}{\mathrm{A}}={10}^2 \mathrm{pm}$.]