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Yuki888 [10]
3 years ago
7

Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an

uranium nuclear fission reaction. Safely disposing of Cs-137 is extremely important due both to this isotope's long half-life, 30.07 years, and it ability to substitute for potassium in biological reactions. A 5.2 mg sample of Cs-137 requires half-lives to decay to 0.65 mg. After 150 years, of Cs-137 remains. After 6 half-lives, of Cs-137 remains. After years, 0.0102 mg of Cs-137 remains. Finally, after approximately 601 years - half-lives, only of the Cs-137 remains.
Physics
1 answer:
vodomira [7]3 years ago
7 0

Answer:

A sample of 5.2 mg  decays to .65 mg or to 1/8 of its original amount.

1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.

3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg

You can get these other numbers similarly:

5.2 / .0102 = 510  requires about 9  half-lives which is 30 * 9 = 270 yrs

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A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
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Answer:

t = 5.59x10⁴ y

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To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

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