Question

Assume that the rate at which radioactive nuclei decay is proportional to the number of such nuclei that are present in a given sample. In a certain sample \(10 \%\) of the original number of radioactive nuclei have undergone disintegration in a period of 100 years. (a) What percentage of the original radioactive nuclei will remain after 1000 years? (b) In how many years will only one-fourth of the original number remain?

Short answer

After 1000 years, approximately 39.97% of the original radioactive nuclei will remain. It will take about 396.85 years for only one-fourth of the original number to remain.

Step-by-step solution

Setup the differential equation

Let N(t) be the number of radioactive nuclei at time t. Since the decay is proportional to the current number of nuclei, we have the following differential equation: \[ \frac{dN}{dt} = -kN, \] where k is the decay constant.

Solve the differential equation

We can solve this first-order linear differential equation by integrating both sides: \[ \int \frac{dN}{N} = -k \int dt. \] On integration, we get: \[ \ln{N} = -kt + C, \] where C is the integration constant. Thus, the solution for N(t) is \[ N(t) = N_0 e^{-kt}, \] where \(N_0 = e^C\) is the initial number of radioactive nuclei.

Determine the decay constant k

We are given that in a period of 100 years, 10% of the original number of radioactive nuclei have undergone disintegration. So after 100 years, we have 90% remaining: \[ N(100) = 0.9N_0. \] Using the decay formula, \[ 0.9N_0 = N_0 e^{-100k}. \] We can now solve for k: \[ 0.9 = e^{-100k} \Rightarrow k = -\frac{\ln{0.9}}{100}. \]

Find the percentage remaining after 1000 years

Using the decay formula with the found k, we can find the percentage remaining after 1000 year since \(N_0\) cancels out: \[ N(1000) = N_0 e^{-1000k} \Rightarrow \frac{N(1000)}{N_0} = e^{-1000(\frac{-\ln{0.9}}{100})}. \] Hence, the percentage remaining after 1000 years is \(100 \times e^{-10\ln{0.9}} \approx 39.97\%\).

Calculating the time required for a quarter to remain

We now want to find the time it takes for only one-fourth of the original number to remain, that is, N(t) = 0.25\(N_0\). Using the decay formula: \[ 0.25N_0 = N_0 e^{-t\frac{-\ln{0.9}}{100}}. \] Now, we can solve for the time t: \[ 0.25 = e^{-t\frac{\ln{0.9}}{100}} \Rightarrow t=-\frac{100\ln{0.25}}{\ln{0.9}} \approx 396.85 \text{ years}. \] To summarize, after 1000 years, there will be approximately 39.97% of the original radioactive nuclei remaining, and it will take about 396.85 years for only one-fourth of the original number to remain.