A certain radioactive material is known to decay at a rate proportional to the amount present. Initially there is a 50kg 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 $50{{\left( \dfrac{9}{10} \right)}^{2}}$
(b) Mass of the material after four hours is $50{{e}^{-0.5\ln 9}}$
(c) Time at which the material has decayed to half of its initial mass (in hours) is $\dfrac{\ln \dfrac{1}{2}}{\left( -\dfrac{1}{2}\ln 0.9 \right)}$
(d) Time at which the material has decayed to half of its initial mass (in hours) is $\dfrac{2\ln \dfrac{1}{2}}{\ln 0.9}$
Answer
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Hint: We are clearly given a point that the decay rate is proportional to the initial amount present. This basically is the law of radioactivity. We could then substitute the value given in the question step by step and thus get the answer. We could do it for time t=4hours and also for amount N=25kg which is half the initial amount.
Formula used:
Law of radioactivity,
$N={{N}_{0}}{{e}^{-\lambda t}}$
Complete answer:
In the question, we are given certain radioactive material that is known to show a decay rate that is proportional to the amount of material present. That is,
$\dfrac{dN}{dt}\alpha N$
$\Rightarrow \dfrac{dN}{dt}=-\lambda N$
$\Rightarrow \dfrac{dN}{N}=-\lambda dt$
$\Rightarrow \ln N=-\lambda t+C$
$\Rightarrow N={{N}_{0}}{{e}^{-\lambda t}}$ …………………………. (1)
We are also given the initial amount of material present at time t=0s as,
$N=50kg$
From (1) we have,
$N=50{{e}^{-\lambda t}}$ ………………………………………. (2)
We are said that 10% of the initial amount is lost after, that is, the amount of material lost would be,
$50-\left( \dfrac{10}{100}\times 50 \right)=50-5=5kg$
So the amount remaining after 2hours would be 45Kg.
Now (1) becomes,
$45=50{{e}^{-2\lambda }}$
$\Rightarrow \dfrac{9}{10}={{e}^{-2\lambda }}$
$\Rightarrow {{e}^{-\lambda }}={{\left( \dfrac{9}{10} \right)}^{\dfrac{1}{2}}}$ …………………………………….. (3)
Now (2) becomes,
$N=50{{\left( \dfrac{9}{10} \right)}^{\dfrac{t}{2}}}$………………………………………. (4)
Put t=4hours in (4),
$N=50{{\left( \dfrac{9}{10} \right)}^{\dfrac{4}{2}}}=50{{\left( \dfrac{9}{10} \right)}^{2}}$………………………………… (5)
When the amount left is half the initial amount, (4) becomes,
$25=50{{\left( \dfrac{9}{10} \right)}^{\dfrac{t}{2}}}$
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}={{\left( \dfrac{9}{10} \right)}^{t}}$
$\Rightarrow 2\ln \left( \dfrac{1}{2} \right)=t\ln \left( \dfrac{9}{10} \right)$
$\therefore t=\dfrac{2\ln \left( \dfrac{1}{2} \right)}{\ln \left( 0.9 \right)}$ ……………………………………………. (6)
Therefore, from (4) and (6), we could conclude that:
(i) Mass of the material after four hours is $50{{\left( \dfrac{9}{10} \right)}^{2}}$
(ii) Time at which the material has decayed to half of its initial mass in hours is $t=\dfrac{2\ln \left( \dfrac{1}{2} \right)}{\ln \left( 0.9 \right)}$
Hence, option A and D are correct.
Note:
In questions like these, where we are given a number of options and we have to find the correct statement among them, chances are there for multiple statements being true. So, we have to check for all the given statements. Thus we have done the calculation for t=4 as well as N=25kg and found multiple answers true.
Formula used:
Law of radioactivity,
$N={{N}_{0}}{{e}^{-\lambda t}}$
Complete answer:
In the question, we are given certain radioactive material that is known to show a decay rate that is proportional to the amount of material present. That is,
$\dfrac{dN}{dt}\alpha N$
$\Rightarrow \dfrac{dN}{dt}=-\lambda N$
$\Rightarrow \dfrac{dN}{N}=-\lambda dt$
$\Rightarrow \ln N=-\lambda t+C$
$\Rightarrow N={{N}_{0}}{{e}^{-\lambda t}}$ …………………………. (1)
We are also given the initial amount of material present at time t=0s as,
$N=50kg$
From (1) we have,
$N=50{{e}^{-\lambda t}}$ ………………………………………. (2)
We are said that 10% of the initial amount is lost after, that is, the amount of material lost would be,
$50-\left( \dfrac{10}{100}\times 50 \right)=50-5=5kg$
So the amount remaining after 2hours would be 45Kg.
Now (1) becomes,
$45=50{{e}^{-2\lambda }}$
$\Rightarrow \dfrac{9}{10}={{e}^{-2\lambda }}$
$\Rightarrow {{e}^{-\lambda }}={{\left( \dfrac{9}{10} \right)}^{\dfrac{1}{2}}}$ …………………………………….. (3)
Now (2) becomes,
$N=50{{\left( \dfrac{9}{10} \right)}^{\dfrac{t}{2}}}$………………………………………. (4)
Put t=4hours in (4),
$N=50{{\left( \dfrac{9}{10} \right)}^{\dfrac{4}{2}}}=50{{\left( \dfrac{9}{10} \right)}^{2}}$………………………………… (5)
When the amount left is half the initial amount, (4) becomes,
$25=50{{\left( \dfrac{9}{10} \right)}^{\dfrac{t}{2}}}$
$\Rightarrow {{\left( \dfrac{1}{2} \right)}^{2}}={{\left( \dfrac{9}{10} \right)}^{t}}$
$\Rightarrow 2\ln \left( \dfrac{1}{2} \right)=t\ln \left( \dfrac{9}{10} \right)$
$\therefore t=\dfrac{2\ln \left( \dfrac{1}{2} \right)}{\ln \left( 0.9 \right)}$ ……………………………………………. (6)
Therefore, from (4) and (6), we could conclude that:
(i) Mass of the material after four hours is $50{{\left( \dfrac{9}{10} \right)}^{2}}$
(ii) Time at which the material has decayed to half of its initial mass in hours is $t=\dfrac{2\ln \left( \dfrac{1}{2} \right)}{\ln \left( 0.9 \right)}$
Hence, option A and D are correct.
Note:
In questions like these, where we are given a number of options and we have to find the correct statement among them, chances are there for multiple statements being true. So, we have to check for all the given statements. Thus we have done the calculation for t=4 as well as N=25kg and found multiple answers true.
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