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Chapter 29: Problem 40

What is the activity in becquerels of \(1.0 \mathrm{kg}\) of $^{238} \mathrm{U} ?$

Short Answer

Expert verified
Question: Calculate the activity in Becquerels (Bq) of 1.0 kg of Uranium-238. Answer: The activity of 1.0 kg of Uranium-238 is 1.244 x 10^7 Becquerels.

Step by step solution

01

Find half-life information

The half-life of Uranium-238, T is given. For Uranium-238, the half-life (\(T_{1/2}\)) is 4.468 x \(10^9\) years.
02

Convert half-life to seconds

To find the activity in becquerels, we need to convert the half-life to seconds. 1 year has 3.1536 x \(10^7\) seconds. So, \(T_{1/2} = 4.468 \times 10^9 \, \text{years} \times 3.1536 \times 10^7 \, \mathrm{s/year} = 1.409 \times 10^{17} \, \mathrm{s}\).
03

Find the decay constant

The decay constant, λ can be calculated from half-life using the formula, \(λ=\frac{\ln{2}}{T_{1/2}}\). So, \(λ = \frac{\ln{2}}{1.409 \times 10^{17} \, \mathrm{s}} = 4.916 \times 10^{-18} \, \mathrm{s}^{-1}\).
04

Calculate the number of Uranium-238 atoms

To calculate the activity, we need the number of Uranium-238 atoms (N) in 1.0 kg of Uranium-238. The atomic mass of Uranium-238 is 238 u. First, find the number of moles of Uranium-238 in 1 kg: \(\text{moles} = \frac{1 \times 10^3 \, \mathrm{g}}{238 \, \mathrm{g/mol}} = 4.202 \times 10^0 \, \mathrm{mol}\). Now, find the number of Uranium-238 atoms, N using Avogadro's number (6.022 x \(10^{23}\,\mathrm{atoms/mol}\)): \(N = 4.202 \times 10^0 \, \mathrm{mol} \times 6.022 \times 10^{23} \, \mathrm{atoms/mol} = 2.530 \times 10^{24} \, \mathrm{atoms}\).
05

Calculate the Activity

The activity A in Bq can be calculated by using the formula \(A = N \cdot λ\). So, \(A = 2.530 \times 10^{24} \,\mathrm{atoms} \times 4.916 \times 10^{-18} \, \mathrm{s}^{-1} = 1.244 \times 10^7 \, \mathrm{Bq}\). Therefore, the activity of \(1.0 \, \mathrm{kg}\) of Uranium-238 is \(1.244 \times 10^7\) Becquerels.

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