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(910)2
B
mass of the material after four hours is 50.e0.5ln9
C
time at which the material has decayed to half of its initial mass (in hours) is (n12)(12n0.9)
D
time at which the material has decayed to half of its initial mass (in hours) is (n2)(12n0.9)

Solution

The correct options are C time at which the material has decayed to half of its initial mass (in hours) is $$\dfrac{(\ell n2)}{(-\frac{1}{2}\ell n0.9)}$$ D mass of the material after four hours is $$50(\dfrac{9}{10})^2$$$$\dfrac{dN}{dt}\propto N$$at $$t = 0 N = 50kg$$and at $$t = 2hr N = 0.9\times 50kg$$$$\dfrac{dN}{dt} = -\lambda N$$$$\Rightarrow \dfrac{dN}{N} = -\lambda dt$$$$\ell nN = -\lambda t+c$$$$t = 0 , N = 50$$$$\ell nN = -\lambda t + \ell n50$$$$N = 50e^{-\lambda t}$$             ......(1)$$t = 2 45=50e^{-\lambda .2}$$$$\dfrac{9}{10} = e^{-2\lambda}$$$$\sqrt{\dfrac{9}{10}} = e^{-\lambda}$$$$\therefore N = 50(\frac{9}{10})^{t/2}$$       ......(2)At $$t = 4$$     $$N = 50(9/10)^2$$When $$N = 25kg 25 = 50(9/10)^{t/2}$$$$\Rightarrow \dfrac{1}{2}=(0.9)^{t/2} t = \dfrac{2\ell n\dfrac{1}{2}}{\ell n0.9}= \dfrac{2\ell n2}{-\dfrac{1}{2}\ell n0.9}$$Mathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More