# 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 ?

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

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.