Math, asked by harshmore790, 10 months ago

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

Answers

Answered by sonuvuce
0

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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