Question

A wooden piece is $11520$ yrs old. What is the fraction of ${}^{14}C$ activity left in the piece?

• A. $0.12$
• B. $0.25$
• C. $0.50$
• D. $0.75$

Answer 1

The correct option is B $0.25$

$\frac{0.693}{t_{1/2}{}^{14}C}=\frac{2.303}{t_{age}}\log \left(\frac{N_0}{N}\right)$
$\frac{0.693}{5760}=\frac{2.303}{11520}\log \left(\frac{N_0}{N}\right)$
$\frac{N}{N_0}=0.25$