Question

The rate, at which the radioactive substance decays, is known to be proportional to the number of nuclei
present at that time in a given sample. In a certain sample, 25% of the original number of radioactive nuclei
have undergone disintegration in a period of 100 years. Find what percentage of the original radioactive
nuclei will remain after 1000 years.​

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Answer 1

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Answer 2

The percentage of the original radioactive nuclei that will remain after 1000 years is 56.31%

Step-by-step explanation:

Let the initial number of radioactive nuclei be $N_0$

If at any period of time, the number of nuclei present is N then

Rate of decay = $-\frac{dN}{dt}$

According to the question, the rate of decay at any given point of time is proportional to the nuclei present

Therefore,

$-\frac{dN}{dt}\propto N$

$\implies -\frac{dN}{dt}=kN$

$\implies -\frac{dN}{N}=kt$

$\implies -\int_{N_0}^N \frac{dN}{N}=k\int_0^t dt$

$\implies -\log N\Bigr|_{N_0}^N=kt\Bigr|_0^t$

$\implies \log {N_0}-\log N=kt$

$\implies \log{\frac{N_0}{N}}=kt$

$\implies \frac{N_0}{N}=e^{kt}$

$\implies N=N_0 e^{-kt}$

Now at t = 100, 25% nuclei have disintigrated i.e. 75% remain

Thus,

$N$ = 75% of $N_0$

$\implies N=0.75N_0$

$\implies N_0 e^{-100k}=0.75N_0$

$\implies e^{-100k}=\frac{3}{4}$

$\implies e^{-k}=(\frac{3}{4})^{1/100}$

$\implies e^k=(\frac{4}{3})^{1/100}$

Number of nuclei remaining after 1000 years

$N=N_0 e^{-1000k}$

$\implies \frac{N}{N_0}=e^{-1000k}$

$\implies (\frac{N}{N_0})^{1/10}=e^{-100k}$

$\implies (\frac{N}{N_0})^{1/10}=\frac{3}{4}$

$\implies \frac{N}{N_0}=(\frac{3}{4})^{10}$

$\implies \frac{N}{N_0}=0.5631$

$\implies \frac{N}{N_0}\%=56.31\%$

Therefore, 56.31 % of the nuclei will remain after 1000 years.

Hope this answer is helpful.

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