#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 28 maths

Given: Radium decomposes at a rate proportional to the quantity of radium present.

To find: The time taken for half the amount of the original amount of radium to decompose.

Hint: The quantity of radium decomposes at the rate of time t i.e. $\frac{d A}{d t} \propto A$  then solve using integration.

Solution: Let A be the quantity of bacteria present in the culture at any time t and initial quantity of bacteria is A0 be the initial quantity of radium.

\begin{aligned} &=\frac{d A}{d t} \propto A \\\\ &=\frac{d A}{d t}=-\lambda A \end{aligned}        [λ is proportional constant]

$=\frac{d A}{A}=-\lambda d t$

Integrating on both sides

$=\int \frac{d A}{A}=-\lambda \int \mathrm{dt}$

$=\log A=-\lambda t+C \ldots(i)$

Now, $A=A_{0}$ when $t=0$

\begin{aligned} &=\log A_{0}=0+C \\\\ &=C=\log A_{0} \end{aligned}

Substituting in equation (i)

\begin{aligned} &=\log A=-\lambda t+\log A_{0} \\\\ &=\log A-\log A_{0}=-\lambda t \\\\ &=\log \frac{A}{A_{0}}=-\lambda t \ldots(i i) \end{aligned}

Given that in 25 years radium decomposes at 1.1%

So,

$\begin{gathered} A=(100-1.1) \%=98.9 \% \\ A=0.989 A_{0}, t=25 \end{gathered}$

Substituting in equation (ii)

\begin{aligned} &=\log \frac{0.989 A_{0}}{A_{0}}=-25 \lambda \\\\ &=\lambda=-\frac{1}{25} \log (0.989) \end{aligned}

Now equation (ii) becomes

$=\log \frac{A}{A_{0}}=\left[-\frac{1}{25} \log (0.989)\right] t$

Now    $A=\frac{1}{2} A_{0}$

\begin{aligned} &=\log \left(\frac{A}{2 A}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=\log \left(\frac{1}{2}\right)=-\frac{1}{25} \log (0.989) t \end{aligned}

\begin{aligned} &=\log \left(2^{-1}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=-1 \log (2)=-\frac{1}{25} \log (0.989) t \end{aligned}

$=t=\frac{\log 2 \times 25}{\log (0.989)}$

$=t=\frac{0.6931 \times 25}{0.01106} \quad[\log 2=0.6931 \text { And } \log (0.989)=0.01106]$

$=t=1567$

Hence, time taken for radium to decay to half the original amount is 1567 years.