Question

The half-life of ${}_{11}^{24}Na$ is $15.0h$. What percentage of it remains after $60h$ ?

• A. $0.00\%$
• B. $3.13\%$
• C. $6.25\%$
• D. $12.5\%$

Answer 1

Answer: (C)

${}_{11}^{24}Na\to {}_{12}^{24}Mg+{}_{-1}^{0}e$
$(\beta )$
$y=$ number of half-life $=\frac{60}{15}=4$
$N=N_0{\left(\frac{1}{2}\right)}^y$
$=100{\left(\frac{1}{2}\right)}^4=\frac{100}{16}=6.25\%$