Question

${15/16}^{th}$ of a radioactive substance disintegrates in 12 days. How much more time will it take for the remaining amount of radioactive substance to reduce to $1/{32}^{th}$ of the reduced amount? What is the half-life period of the radioactive substance?

• A. 40 days, 3 days
• B. 30 days, 3 days
• C. 15 days, 3 days
• D. 5 days, 2 days

Answer 1

Answer: (C)

15 days, 3 days

$\begin{array}{c}\text{Let there be}16\text{g of substance}\\ \text{After}12\text{days,}1g\text{remains.}\\ \text{we know,}N=N_0{e}^{-\lambda t}\end{array}$

$\begin{array}{cc}1& =16e^{-\lambda \left(12\right)}\\ \Rightarrow \ln 16& =12\lambda \\ \text{or,}\lambda & =\frac{\ln 2}{3}.\end{array}$

$\begin{array}{c}\text{Now, for substance to reduce to}{\frac{1}{32}}^{th}\text{of}1g\\ \Rightarrow \frac{1}{32}=1e^{-\frac{\ln 2}{3}\times t}\\ \Rightarrow \ln 32=\frac{\ln 2}{3}t\\ \Rightarrow t=15\text{days}\end{array}$

$\begin{array}{cc}\text{Hall-life:}{t}_{1/2}& =\frac{\ln 2}{\lambda }\\ & =\frac{\ln 2}{\ln 2}\times 3\\ & =3\text{days}\end{array}$