Answer :
Final answer:
The half-life of cesium-137 is 30 years. It would take approximately 270 years to reduce a 1 kg sample of cesium-137 to about 1 g.
Explanation:
The half-life of cesium-137 is 30 years. This means that every 30 years, the amount of cesium-137 in a sample is halved. To find out approximately how much time will be required to reduce a 1 kg sample to about 1 g, we can use the following calculations:
- After 30 years, the sample will have halved to 0.5 kg
- After another 30 years (60 years in total), the sample will have halved to 0.25 kg
- After another 30 years (90 years in total), the sample will have halved to 0.125 kg
- After another 30 years (120 years in total), the sample will have halved to 0.0625 kg
- After another 30 years (150 years in total), the sample will have halved to 0.03125 kg
- After another 30 years (180 years in total), the sample will have halved to 0.015625 kg
- After another 30 years (210 years in total), the sample will have halved to 0.0078125 kg
- After another 30 years (240 years in total), the sample will have halved to 0.00390625 kg
- After another 30 years (270 years in total), the sample will have halved to 0.001953125 kg (approximately 1 g)
So it would take approximately 270 years to reduce a 1 kg sample of cesium-137 to about 1 g.
Learn more about Half-life of cesium-137 here:
https://brainly.com/question/32919758
#SPJ11
Answer:
The time required is [tex]t = 299 \ years[/tex]
Explanation:
From the question we are told that
The half-life of cesium-137 is [tex]t_{h} = 30 \ years[/tex]
The mass of the sample is [tex]A_o= 1 kg = 1000 \ g[/tex]
The reduced mass of the sample is [tex]A = 1 \ g[/tex]
The remaining sample can be mathematically calculated as
[tex]A = A_o e ^{-\lambda t }[/tex]
Where [tex]\lambda[/tex] is the decay constant which is mathematically represented as
[tex]\lambda = \frac{t_h}{ln 2 }[/tex]
substituting value
[tex]\lambda = \frac{0.693}{30}[/tex]
[tex]\lambda = 0.0231[/tex]
and t is the time taken
So the above equation becomes
[tex]A = A_o e ^{-0.0231 t }[/tex]
=> [tex]ln [\frac{1}{1000} ] = {-0.0231 t }[/tex]
[tex]-6.907755 = {-0.0231 t }[/tex]
[tex]t = 299 \ years[/tex]