Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

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 $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App