Question

A sample contains 0.1 gram - atom of radioactive isotope ${}_Z^A\text{X}(t_{1/2}=5days)$. How many number of atoms will decay during eleventh day? $[N_A=Avogadr{o}^{\prime }snumber]$

• A. $0.1(-e^{\frac{0.693\times 11}{5}}+e^{\frac{0.693\times 10}{5}})$
• B. $0.1(-e^{\frac{0.693\times 11}{6}}+e^{\frac{0.693\times 10}{5}})$
• C. $0.1(-e^{-\frac{0.693\times 11}{5}}+e^{-\frac{0.693\times 10}{5}})N_A$
• D. $0.1(e^{-\frac{0.693\times 11}{5}}+e^{\frac{0.693\times 10}{5}})N_A$

Answer 1

The correct option is C $0.1(-e^{-\frac{0.693\times 11}{5}}+e^{-\frac{0.693\times 10}{5}})N_A$

$\lambda =\frac{0.693}{t_{1/2}}=\frac{0.693}{5}$
$a-x=a{e}^{-\lambda t}=$
For $t=11days$
$0.1-x=0.1N_A{e}^{-\frac{0.693\times 11}{5}}=$......(1)
For $t=10days$
$0.1-x=0.1N_A{e}^{-\frac{0.693\times 10}{5}}=$......(2)
From equation (1) and (2), the number of atoms will decay during eleventh day is $0.1(-e^{-\frac{0.693\times 11}{5}}+e^{-\frac{0.693\times 10}{5}})N_A$