To solve the question of what percentage of uranium-238 will remain after 2.5 billion years, given that the half-life of uranium-238 is 4.2 billion years, we'll use the half-life formula for radioactive decay. The formula is:
[tex]\[ A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{H}} \][/tex]
where:
- [tex]\( A \)[/tex] is the remaining quantity of the substance after time [tex]\( t \)[/tex].
- [tex]\( A_0 \)[/tex] is the initial quantity of the substance.
- [tex]\( t \)[/tex] is the time that has passed.
- [tex]\( H \)[/tex] is the half-life of the substance.
In this context, we are interested in finding the percentage of the remaining uranium-238 ([tex]\( A \)[/tex]) after 2.5 billion years.
Let's break it down step-by-step:
1. Identify the known values:
- Half-life ([tex]\( H \)[/tex]) of uranium-238 = 4.2 billion years
- Time passed ([tex]\( t \)[/tex]) = 2.5 billion years
2. Substitute the values into the formula:
[tex]\[ A = A_0 \left(\frac{1}{2}\right)^{\frac{t}{H}} \][/tex]
[tex]\[ A = A_0 \left(\frac{1}{2}\right)^{\frac{2.5}{4.2}} \][/tex]
3. Simplify the exponent:
Calculate [tex]\(\frac{2.5}{4.2} \approx 0.5952\)[/tex].
4. Evaluate the expression [tex]\(\left(\frac{1}{2}\right)^{0.5952}\)[/tex]:
When you raise [tex]\(\frac{1}{2}\)[/tex] to the power of approximately 0.5952, you get approximately 0.661935.
5. Convert the remaining fraction to a percentage:
The remaining percentage of uranium-238 is then approximately:
[tex]\[ 0.661935 \times 100 = 66.1935\% \][/tex]
So, after 2.5 billion years, approximately 66.1935% of uranium-238 will remain, given its half-life of 4.2 billion years.
### Self-Evaluation
I rate my work a 5. This problem required understanding the application of the half-life formula in the context of radioactive decay. Through solving this, I reinforced my knowledge of logarithmic and exponential functions and how they apply to real-world scenarios. The most challenging part was ensuring accuracy in the simplification and exponentiation steps, but it helped solidify my problem-solving skills in this area.
