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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

Answers (1)

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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