Question

You have $0.1g$ atom of a radioactive isotope ${}_Z^{A}X$ (half-life $=5$ days). How many atoms will decay during the $11$th day?

Answer 1

Amount of radioactive substance $=0.1g$ atom
So, $N_0=0.1\times$ Avogadro's number
$=0.1\times 6.02\times {10}^{23}$
$=6.02\times {10}^{22}$ atoms
Number of atoms after $5$ days $=\frac{6.02\times {10}^{22}}{2}=3.01\times {10}^{22}$
Number of atoms after $10$ days $=\frac{3.01\times {10}^{22}}{2}=1.505\times {10}^{22}$
Let the number of atoms left after $11$ days be $N$.
We know that,
$t=\frac{2.303}{\lambda }{\log }_{10}\frac{N_0}{N}$
Given, $t=11$, $\lambda =\frac{0.693}{5}$, $N_0=6.02\times {10}^{22}$
So, $11=\frac{2.303\times 5}{0.693}{\log }_{10}\frac{6.02\times {10}^{22}}{N}$
or ${\log }_{10}\frac{6.02\times {10}^{22}}{N}=\frac{11\times 0.693}{2.303\times 5}=0.6620$
$\frac{6.02\times {10}^{22}}{N}=Antilog0.6620=4.592$
So, $N=\frac{6.02}{4.592}\times {10}^{22}=1.3109\times {10}^{22}$
Atoms decayed during $11$th day $=[1.5050\times {10}^{22}-1.3109\times {10}^{22}]$
$=0.1941\times {10}^{22}$
$=1.941\times {10}^{21}$