# The half-life of cesium-137 is 30 years. Assuming you start with a 100 mg sample of cesium-137. A) How much cesium-137 will be left after 2 half-lives? B) How long will it take for the 100 mg of...

The half-life of cesium-137 is 30 years. Assuming you start with a 100 mg sample of cesium-137.

A) How much cesium-137 will be left after 2 half-lives?

B) How long will it take for the 100 mg of cesium-137 to decay to 6.25 mg (1/16 the original amount)?

Thanks in advance.

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### 1 Answer

The relation between the initial number of existent nuclei `N_0` , the number of remaining nuclei at time t `N(t)` , and half life `t_(1/2)` is

`N(t) = N_0*2^(-t/t_(1/2))`

and since in one mole of any substance enters the same number of atoms (or molecules) - Avogadro number, then we have the same equation for masses (initial and at time t)

`m(t) = m_0*2^(-t/t_(1/2))`

A)

Thus if `t =2*t_(1/2)` we have

`m(60 years) =m_0*2^(-2) =m_0/4 =100/4 = 25 mg`

B)

In the case `m_0 =100 mg ` and `m(t) =6.25 mg =m_0/16` we can write

`m_0/16 =m_0*2^(-t/t_(1/2))`

`m_0*2^-4 =m_0*2^(-t/t_(1/2))`

`t/t_(1/2) =4`

or equivalent `t =4*t_(1/2) =4*30 =120 years`

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Thank you very much