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

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

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Answer: 0.04%

Given: Half like of radium=1590 years.

To find: We have to find the percentage of radium decaying in one year.

Hint:Take \frac{d A}{d t}=-\lambda t   and integrate it to find rate of  \lambda

Solution: Given that

        \begin{aligned} &\frac{d A}{d t} \propto A \\ &=\frac{d A}{d t}=-\lambda A \end{aligned}               [Where ?the constant of proportionality and minus sign is indicates ]

        =\frac{d A}{A}=-\lambda d t            [that A decrease with increase in t]

Integrating on both sides

        \begin{aligned} &=\int \frac{d A}{A}=-\lambda \int d t\\\\ &=\log A=-\lambda t+C \end{aligned}    ........(i)

Since initial amount of radium is A0 then

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

Putting value of C in equation (i) we get

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

Given that its half-life is 1590 years so we put A=\frac{1}{2} A_{0}   and t=1590

Hence, equation (ii) becomes

        \begin{aligned} &=>\log \left(\frac{A}{A_{0}}\right)=-\lambda t \\ &=> \log \left(\frac{A_{0}}{2 A_{0}}\right)=-\lambda \times 1590 \end{aligned}

        \begin{aligned} &=>\log \left(\frac{1}{2}\right)=-\lambda(1590) \\\\ &=>-\log 2=-\lambda \times 1590 \end{aligned}

Taking minus sign on both sides,

        \begin{aligned} &=>\log 2=1590 \lambda \\\\ &=>\lambda=\frac{\log 2}{1590} \end{aligned}

Put =\lambda=\frac{\log 2}{1590}   in equation (ii) we get

        \begin{aligned} &=>\log \left(\frac{A}{A_{0}}\right)=-\left(\frac{\log 2}{1590}\right) t \\\\ &=>\frac{A}{A_{0}}=e^{-\left(\frac{\log 2}{1590}\right) t} \ldots(i i i) \end{aligned}

Putting t=1 in (iii) to find the amount of radium after one year, we get

        \begin{aligned} &=\frac{A}{A_{0}}=e^{-\frac{\log 2}{1590}} \\\\ &=>\frac{A}{A_{0}}=0.9996 \\\\ &=>A=A_{0} 0.9996 \end{aligned}

Percentage amount disappeared in one year

        =\frac{A_{0}-A}{A_{0}} \times 100 \\

        =\frac{A_{0}-A_{0} 0.9996}{A_{0}} \times 100

        =\frac{A_{0}(1-0.9996)}{A_{0}} \times 100 \\

        =(1-0.9996) \times 100 \\

        =0.0004 \times 100 \\

        =0.04 \%

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