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

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