Explanation:
Let the age to be found in years is y.
Hence, [tex](2480 disintegrations) \times \frac{\frac{1.0g}{0.250g}}{1.84 \times 10^{4} disintegrations}[/tex]
= (\frac{1}{2})^{\frac{y}{5730yr}}
Solve for y as follows.
0.53913 = (\frac{1}{2})^{\frac{y}{5730yr}}
Now, taking log on both the sides as follows
log 0.53913 = [tex](\frac{z}{5730}) log \frac{1}{2} [/tex]
[tex]\frac{log 0.53913}{log (1/2)}[/tex] = [tex]\frac{z}{5730}[/tex]
z = [tex]\frac{5730 \times log 0.53913}{log (1/2)}[/tex]
= 5107 years
Thus, we can conclude that the time since death is 5107 years.
