Respuesta :
Answer:
From what I can see of the problem, you CANNOT solve for the half-life of U 235 AND then solve for the length of time to determine the 20% decay.
U-235 half-life is 704,000,000 years. (Wikipedia)
The elapsed time formula = half-life * [log (Beginning Amount / ending amount) / log 2]
elapsed time = 7.04 x 10^8 * [log (100 % / 80%) / log 2]
elapsed time =7.04 x 10^8 * [log (1.25) / .30103]
elapsed time =7.04 x 10^8 * [0.096910 / .30103]
elapsed time = 7.04 x 10^8 * 0.321928047
elapsed time = 226,637,000 years
Step-by-step explanation:
From the exponential equation, we have that:
- For Uranium 235, [tex]T = 700000000[/tex].
- Solving the model considering A(T) = 0.5A(0), it is proved that [tex]K = \frac{\ln{0.5}}{T}[/tex].
- It will take 225,349,666 years before 20% of the sample decays.
-----------------
- The half-life of Uranium 235 is of 700 million years, thus [tex]T = 700000000[/tex]
The exponential equation that models the amount remaining after t years is given by:
[tex]A(t) = A(0)e^{Kt}[/tex]
Solving for the half-life, that is, [tex]A(T) = 0.5A(0)[/tex], we find k.
[tex]0.5A(0) = A(0)e^{Kt}[/tex]
[tex]e^{KT} = 0.5[/tex]
[tex]\ln{e^{KT}} = \ln{0.5}[/tex]
[tex]K = \frac{\ln{0.5}}{T}[/tex]
Here is the proof.
Considering [tex]T = 700000000[/tex]
[tex]K = \frac{\ln{0.5}}{700000000} = 9.9 \times 10^{-10}[/tex]
Thus:
[tex]A(t) = A(0)e^{-9.9 \times 10^{-10}t}[/tex]
How long will it take before twenty percent of our 1,000-gram sample of uranium-235 has decayed?
This is t for which A(t) = 0.8A(0), as 100% - 20% = 80% remaining.
[tex]A(t) = A(0)e^{-9.9 \times 10^{-10}t}[/tex]
[tex]0.8A(0) = A(0)e^{-9.9 \times 10^{-10}t}[/tex]
[tex]e^{-9.9 \times 10^{-10}t} = 0.8[/tex]
[tex]\ln{e^{-9.9 \times 10^{-10}t}} = \ln{0.8}[/tex]
[tex]9.9 \times 10^{-10}t = -\ln{0.8}[/tex]
[tex]t = -\frac{\ln{0.8}}{9.9 \times 10^{-10}}[/tex]
[tex]t = 225349666[/tex]
It will take 225,349,666 years before 20% of the sample decays.
A similar problem is given at https://brainly.com/question/24829007
