Ask Questions, Get Answers

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

A radioactive substance disintegrates at a rate proportional to its mass. When its mass is $10 $mgm, the rate of disintegration is $0.051$ mgm per day. How long will it take for the mass to be reduced from $10$mgm to $5$mgm.$ [\log_{e}$$2=0.6931]$

Can you answer this question?

1 Answer

0 votes
  • First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
  • It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
  • The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$
Step 1:
Let $x$ be the mass of the substance at time $t$ in days
When $t=0,x_0=10mg$ and $\large\frac{dx}{dt}$=0.051mg/day
Step 2:
Now $\large\frac{dx}{dt}\propto x$
It is negative because there is decay
When $t=0$
$\Rightarrow k=-0.0051$
Step 3:
$\therefore \large\frac{dx}{dt}$$=-0.0051x$
Step 4:
Integrating on both sides we get,
$\int \large\frac{dx}{x}$$=0.0051t+\log c$
$\log x=-0.0051t+\log c$
$\log \large\frac{x}{c}$$=-0.0051t$
$x=ce^{\large -0.0051t}$
Step 5:
When $t=0,x_0=10=c$
$\therefore x=10e^{\large -0.0051t}$
Step 6:
When $x=5mg$,to find $t$
$5=10e^{\large 0.0051t}$
$\therefore \large\frac{1}{2}=e^{0.0051t}$
Or $\log _e\large\frac{1}{2}$$=-0.0051t$
$-\log _e2=-0.0051t$
$t=\large\frac{\log _e2}{0.0051}$
$\log _e2=0.6931$
$\therefore t=\large\frac{0.6931}{0.0051}$
Therefore the mass is reduced from 10mg to 5mg in 136 days approximately.
answered Sep 8, 2013 by sreemathi.v

Related questions

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