Question

A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is 50 kg of the material present and after two hours it is observed that the material has lost 10%of its original mass, then the

• A. mass of the material after four hours is $50(\frac{9}{10}{)}^2$
• B. mass of the material after four hours is $50.e^{-0.5ln9}$
• C. time at which the material has decayed to half of its initial mass (in hours) is $\frac{\left(\ell n\frac{1}{2}\right)}{(-\frac{1}{2}\ell n0.9)}$
• D. time at which the material has decayed to half of its initial mass (in hours) is $\frac{\left(\ell n2\right)}{(-\frac{1}{2}\ell n0.9)}$

Answer 1

Answer: (A), (D)

time at which the material has decayed to half of its initial mass (in hours) is $\frac{\left(\ell n2\right)}{(-\frac{1}{2}\ell n0.9)}$
mass of the material after four hours is $50(\frac{9}{10}{)}^2$

$\frac{dN}{dt}\propto N$
at $t=0N=50kg$
and at $t=2hrN=0.9\times 50kg$
$\frac{dN}{dt}=-\lambda N$

$\Rightarrow \frac{dN}{N}=-\lambda dt$

$\ell nN=-\lambda t+c$

$t=0,N=50$

$\ell nN=-\lambda t+\ell n50$

$N=50e^{-\lambda t}$ ......(1)

$t=245=50e^{-\lambda .2}$

$\frac{9}{10}=e^{-2\lambda }$

$\sqrt{\frac{9}{10}}=e^{-\lambda }$

$\therefore N=50(\frac{9}{10}{)}^{t/2}$ ......(2)
At $t=4$
$N=50(9/10{)}^2$
When $N=25kg25=50(9/10{)}^{t/2}$
$\Rightarrow \frac{1}{2}=(0.9{)}^{t/2}t=\frac{2\ell n\frac{1}{2}}{\ell n0.9}=\frac{2\ell n2}{-\frac{1}{2}\ell n0.9}$