Question

0.1 g-atom of radioactive isotope ${}_Z^A\text{X}$ (half life 5 day) are taken. The number of atoms will decay during the eleventh day is :

• A. $1.93\times {10}^{21}atoms$
• B. $1.55\times {10}^{21}atoms$
• C. $2.64\times {10}^{21}atoms$
• D. None of these

Answer 1

The correct option is A $1.93\times {10}^{21}atoms$

Given, The initial amount of radioisotope $N_0=0.1$ g-atom and the half life period $t_{1/2}=5$ day decay time $t=10day$
The integrated rate law expression for the first order decay process is
$\lambda =\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log \frac{N_0}{N}$
or $\frac{0.693}{5}=\frac{2.303}{10}\log \frac{0.1}{N_{10}}$
Amount left after 10 days $=\left(N_{10}\right)=0.0250$ g-atom
Similarly, if the decay period $t=11$ day
$\lambda =\frac{0.693}{t_{1/2}}=\frac{2.303}{t}\log \frac{N_0}{N_{11}}$ or $\frac{0.693}{5}=\frac{2.303}{10}\log \frac{0.1}{N_{11}}$
Amount left after 11 days $\left(N_{11}\right)=0.0218$ g-atom
Amount decayed in 11th day $=0.0250-0.0218$
$=3.2\times {10}^{-3}$g- atom
$=3.2\times {10}^{-3}\times 6.023\times {10}^{23}$ atoms
$=1.93\times {10}^{21}$ atoms