Question

The analysis of a rock shows the relative number of $U^{238}$ and $P{b}^{206}$ atoms (Pb/U = 0.25). The age of rock will be :

• A. $\frac{2.303}{0.693}\times 4.5\times {10}^{9}log1.25$
• B. $\frac{2.303}{0.693}\times 4.5\times {10}^{9}log0.25$
• C. $\frac{2.303}{0.693}\times 4.5\times {10}^{9}log4$
• D. $\frac{2.303}{4.5\times {10}^9}\times 0.693log4$

Answer 1

The correct option is A $\frac{2.303}{0.693}\times 4.5\times {10}^{9}log1.25$

$\frac{0.693}{t_{1/2}\left(U^{238}\right)}=\frac{2.303}{t_{age}}log\left(\frac{N_0}{N}\right)$
$\frac{0.693}{4.5\times {10}^9}=\frac{2.303}{t_{age}}log1.25$
$\therefore t_{age}=\frac{2.303\times 4.5\times {10}^9}{0.693}log1.25$