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A ratio active substance dis integrals at a rate proportional to amount of substance present. $50 \%$ of amount disintegrates in $1000$ years. Approximately what $\%$ of substance will disintegrate in $50$ years ?

$(a)\;(1-(\frac{1}{2}) ^{1/20}) \times 100 \\ (b)\;(1-(\frac{3}{4}) ^{1/20}) \times 100 \\ (c)\;(\frac{1}{2}) ^{1/20} \times 100 \\ (d)\;(1-(\frac{3}{2}) ^{1/20}) \times 100 $

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Let initially $N_0$ no of particles of substance present.
Then $\large\frac{dN}{dt} $$=KN$
So $\log N \bigg]_{N_0}^{N/2}=Kt\bigg]_0^{1000}$
$\log \frac{1}{2} =1000K$
$\log N \bigg]_{N_0}^N=Kt\bigg]_0^{50}$
$\log \large\frac{N}{N_0}=\large\frac{50}{1000} \times $$ \log (1/2)$
$\log \large\frac{N}{N_0}=\large\frac{1}{20} \times $$ \log (1/2)$
$\log \large\frac{N}{N_0}=\bigg(\large\frac{1}{2}\bigg)^{\Large\frac{1}{20}}$
No of particles at t time =N
No of particles dis integrate $=N_0-N$
$\%= \large\frac{N_0-N}{N_0} $$\times 100 =\bigg(1- \bigg( \large\frac{1}{2}\bigg)^{1/20}\bigg) \times 100$
Hence a is the correct answer.
answered Feb 6, 2014 by meena.p

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