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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]$

1 Answer

  • 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

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