We appreciate your visit to Cesium 137 was a radioactive chemical released at the Chernobyl nuclear reactor in April 1986 The estimated amount of cesium 137 released was 3980 grams. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
It will take approximately 196.41 years for the cesium-137 released during the Chernobyl disaster to decay to 31.09 grams, given its half-life of 30.22 years. This can be calculated using the formula for radioactive decay. The initial amount released was 3980 grams.
Cesium-137 was a radioactive chemical released at the Chernobyl nuclear reactor in April 1986. The estimated amount of Cesium-137 released was 3980 grams. The half-life of cesium-137 is 30.22 years. How long will it take for 31.09 grams to remain in the environment?
To determine how long it will take for the remaining mass to reach 31.09 grams, we can use the formula:
N(t) = N0 × [tex](1/2)^(t/T)[/tex]
Where:
N(t) is the remaining amount of the substance after time t. N0 is the initial amount of the substance. T is the half-life of the substance. t is the time elapsed.
Given:
N0 = 3980 grams N(t) = 31.09 grams T = 30.22 years
Rearranging the formula to solve for t:
t = T × (log(N(t)/N0) ÷ log(1/2))
Substituting the values:
t = 30.22 × (log(31.09/3980) ÷ log(1/2))
Calculating the values:
t = 30.22 × (log(0.00781) ÷ log(0.5)) t ≈ 196.41 years
It will take approximately 196.41 years for the cesium-137 to decay to 31.09 grams.
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It will take approximately 212.07 years for the amount of Cesium-137 to decay from 3980 grams to 31.09 grams.
To determine how long it will take for 31.09 grams of Cesium-137 to remain in the environment, we can use the half-life formula for exponential decay:
[tex]\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \][/tex]
where:
N(t) is the amount of the substance that remains after time (t),
[tex]N_0[/tex] is the initial amount of the substance,
[tex]T_{1/2}[/tex] is the half-life of the substance, and
t is the time elapsed.
Given:
- [tex]\( N_0 = 3980 \)[/tex] grams (initial amount of Cesium-137),
- [tex]\( T_{1/2} = 30.22 \)[/tex] years (half-life of Cesium-137),
- [tex]\( N(t) = 31.09 \)[/tex] grams (amount of Cesium-137 remaining after time (t).
We want to find (t) when [tex]\( N(t) = 31.09 \)[/tex] grams. Rearranging the half-life formula to solve for (t), we get:
[tex]\[ t = T_{1/2} \frac{\log\left(\frac{N(t)}{N_0}\right)}{\log\left(\frac{1}{2}\right)} \][/tex]
Now, plug in the given values:
[tex]\[ t = 30.22 \times \frac{\log\left(\frac{31.09}{3980}\right)}{\log\left(\frac{1}{2}\right)} \][/tex]
Using a calculator to find the value of (t):
[tex]\[ t = 30.22 \times \frac{\log\left(0.007796\right)}{\log\left(0.5\right)} \][/tex]
[tex]\[ t \approx 30.22 \times \frac{-2.1139}{\log\left(0.5\right)} \][/tex]
[tex]\[ t \approx 30.22 \times \frac{-2.1139}{-0.3010} \][/tex]
[tex]\[ t \approx 30.22 \times 7.0226 \][/tex]
[tex]\[ t \approx 212.07 \][/tex]
