Question

The half-life of ${}_{27}^{60}Co$ is 5.3 years. How much of 20 g of ${}_{27}^{60}Co$ will remain radioactive after 21.2 years?

• A. 10 g
• B. 1.25 g
• C. 2.5 g
• D. 3.0 g

Answer 1

Answer: (B)

1.25 g

The half-life $t_{1/2}=5.3$ years.

The decay constant $\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{5.3}=0.13076$ /year
$a=20$ g

$a-x=$ ??

$t=21.2$ year

$t=\frac{2.303}{\lambda }log\frac{a}{a-x}$

$21.2=\frac{2.303}{0.13076}log\frac{20}{a-x}$

$1.203=log\frac{20}{a-x}$

$15.98=\frac{20}{a-x}$

$a-x=\frac{20}{15.98}=1.25$ g

Hence, 1.25 g of $Co-60$ will remain radioactive after 21.2 years.