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If the half-life of cesium-137 is 30 years, approximately how much time will be required to reduce a 1 kg sample to about 1 g?

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:

  1. After 30 years, the sample will have halved to 0.5 kg
  2. After another 30 years (60 years in total), the sample will have halved to 0.25 kg
  3. After another 30 years (90 years in total), the sample will have halved to 0.125 kg
  4. After another 30 years (120 years in total), the sample will have halved to 0.0625 kg
  5. After another 30 years (150 years in total), the sample will have halved to 0.03125 kg
  6. After another 30 years (180 years in total), the sample will have halved to 0.015625 kg
  7. After another 30 years (210 years in total), the sample will have halved to 0.0078125 kg
  8. After another 30 years (240 years in total), the sample will have halved to 0.00390625 kg
  9. 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.

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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]

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