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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, 位 can be calculated from half-life using the formula, \(位=\frac{\ln{2}}{T_{1/2}}\). So, \(位 = \frac{\ln{2}}{1.409 \times 10^{17} \, \mathrm{s}} = 4.916 \times 10^{-18} \, \mathrm{s}^{-1}\).
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 \(A = N \cdot 位\). 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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Most popular questions from this chapter

In this problem, you will verify the statement (in Section 29.4) that the \(^{14} \mathrm{C}\) activity in a living sample is \(0.25 \mathrm{Bq}\) per gram of carbon. (a) What is the decay constant \(\lambda\) for \(^{14} \mathrm{C} ?\) (b) How many \(^{14} \mathrm{C}\) atoms are in \(1.00 \mathrm{g}\) of carbon? One mole of carbon atoms has a mass of \(12.011 \mathrm{g},\) and the relative abundance of \(^{14} \mathrm{C}\) is \(1.3 \times 10^{-12} .\) (c) Using your results from parts (a) and (b), calculate the \(^{14} \mathrm{C}\) activity per gram of carbon in a living sample.
The activity of a sample containing radioactive \(^{108} \mathrm{Ag}\) is $6.4 \times 10^{4}\( Bq. Exactly 12 min later, the activity is \)2.0 \times 10^{3} \mathrm{Bq} .\( Calculate the half-life of \)^{108} \mathrm{Ag} .$
Suppose that a radioactive sample contains equal numbers of two radioactive nuclides \(A\) and \(B\) at \(t=0 .\) A has a half-life of \(3.0 \mathrm{h},\) while \(\mathrm{B}\) has a half-life of \(12.0 \mathrm{h}\) Find the ratio of the decay rates or activities \(R_{\mathrm{A}} / R_{\mathrm{B}}\) at (a) \(t=0,\) (b) \(t=12.0 \mathrm{h},\) and \((\mathrm{c}) t=24.0 \mathrm{h}\).
A space rock contains \(3.00 \mathrm{~g}\) of \({ }_{62}^{147} \mathrm{Sm}\) and \(0.150 \mathrm{~g}\) of ${ }_{60}^{143} \mathrm{Nd} .{ }_{62}^{147} \mathrm{Sm} \alpha\( decays to \){ }_{60}^{143} \mathrm{Nd}\( with a half-life of \)1.06 \times 10^{11} \mathrm{yr}\(. If the rock originally contained no \){ }_{60}^{143} \mathrm{Nd}$ how old is it?
Using a mass spectrometer, the mass of the \({ }_{92}^{238} \mathrm{U}^{+}\) ion is found to be 238.05024 u. (a) Use this result to calculate the mass of the \({ }_{92}^{238} \mathrm{U}\) nucleus. (b) Now find the binding energy of the \({ }_{92}^{238} \mathrm{U}\) nucleus.
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