Question

The half-line 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. 10g
• B. 1.25g
• C. 2.5g
• D. 3.0g

Answer 1

The correct option is B 1.25g

Using the formula, $N=N_0{\left(\frac{1}{2}\right)}^n$

where, $N_0=$ initial amount of radioactive substance

$N=$ Amount of substance left after 'n' half lives

No. of half lives $\left(n\right)=\frac{\text{Total time (t)}}{\text{Half lifeperiod}}$

$n=\frac{21.2years}{5.3years}$

$n=4$

so, $N=20{\left(\frac{1}{2}\right)}^4=1.25g$