Questions: If uranium-238 has a half-life of 3.69 × 10^9 years, how many millennium (1000 years) will it take for 4 grams of depleted uranium to be depleted by 0.9 grams of its original mass? (Round to the nearest whole number)

Solution Steps

We need to determine how many millennia it will take for 4 grams of uranium-238 to lose 0.9 grams of its original mass. This means we want to find the time it takes for the mass to reduce to 3.1 grams (4 grams - 0.9 grams).

The decay of a radioactive substance can be described by the formula:

\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]

where:

• \(N(t)\) is the remaining quantity of the substance after time \(t\),
• \(N_0\) is the initial quantity of the substance,
• \(T_{1/2}\) is the half-life of the substance,
• \(t\) is the time elapsed.

Given:

• \(N_0 = 4\) grams,
• \(N(t) = 3.1\) grams,
• \(T_{1/2} = 3.69 \times 10^9\) years.

We need to solve for \(t\) in the equation:

\[ 3.1 = 4 \left(\frac{1}{2}\right)^{\frac{t}{3.69 \times 10^9}} \]

First, divide both sides by 4:

\[ \frac{3.1}{4} = \left(\frac{1}{2}\right)^{\frac{t}{3.69 \times 10^9}} \]

Calculate \(\frac{3.1}{4} = 0.775\).

Take the natural logarithm of both sides:

\[ \ln(0.775) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{3.69 \times 10^9}}\right) \]

Using the property of logarithms, this becomes:

\[ \ln(0.775) = \frac{t}{3.69 \times 10^9} \ln\left(\frac{1}{2}\right) \]

Solve for \(t\):

\[ t = \frac{\ln(0.775)}{\ln(0.5)} \times 3.69 \times 10^9 \]

Calculate the values:

\[ \ln(0.775) \approx -0.2549, \quad \ln(0.5) \approx -0.6931 \]

\[ t \approx \frac{-0.2549}{-0.6931} \times 3.69 \times 10^9 \approx 1.356 \times 10^9 \text{ years} \]

Convert the time from years to millennia (1000 years):

\[ t \approx \frac{1.356 \times 10^9}{1000} \approx 1.356 \times 10^6 \text{ millennia} \]

Round to the nearest whole number:

\[ t \approx 1356 \text{ millennia} \]

Final Answer