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

Hint:The elements which release some particles to get stability are called radioactive elements. The time required to decay half of its original concentration by a radioactive element is called half-time. The symbol to denote the half-life of a radioactive element is ${{t}_{{}^{1}/{}_{2}}}$ .
\[t=\frac{2.303}{\lambda }\log \left( \frac{a}{a-x} \right)\]
Here t = total time in years
$\lambda $ = decay constant = $\frac{0.639}{{{t}_{{}^{1}/{}_{2}}}}$
a = total amount =
a-x = amount after decay

Complete step-by-step answer:- In the given to calculate the amount of ${}_{27}^{60}Co$ left after 21.2 years having a half-life of 5.3 years.
- The formula to calculate the amount of substance left is as follows.
\[t=\frac{2.303}{\lambda }\log \left( \frac{a}{a-x} \right)\]
Here t = total time in years = 21.2
$\lambda $ = decay constant = $\frac{0.639}{{{t}_{{}^{1}/{}_{2}}}}$
a = total amount = 20
a-x = amount after decay
- First we have to calculate the decay constant of the radioactive element and it is as follows.
$\lambda =\frac{0.639}{{{t}_{{}^{1}/{}_{2}}}}=\frac{0.639}{5.3}=0.13076per\text{ }year$
- Now substitute all the known values in the above formula to get the amount of substance left after 21.2 years.
- Therefore the amount of radioactive substance which is left after 21.2 years is 1.25 g.
\[ t=\frac{2.303}{\lambda }\log \left( \frac{a}{a-x} \right) \\
\Rightarrow 21.2=\frac{2.303}{0.13076}\log \left( \frac{20}{a-x} \right) \\
\Rightarrow 1.230=\log \left( \frac{20}{a-x} \right) \\
\Rightarrow 15.98=\frac{20}{a-x} \\
\therefore a-x=1.25g \]

Note:The radioactive elements are the elements which do not have stability due to the presence of high energy. to decrease their energy they are going to emit some particles in the form of radiation and get stability. The emission of particles is called decay.