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Chapter 19: Problem 60

(a) What is the activity, in millicuries, of a \(0.500-\mathrm{g}\) sample of \({ }_{93}^{237} \mathrm{~Np} ?\) (This isotope decays by \(\alpha\) -particle emission and has a half-life of \(2.20 \times 10^{6}\) yr. (b) Write a balanced nuclear equation for the decay of \({ }_{93}^{237} \mathrm{~Np}\).

Short Answer

Expert verified
The activity of a 0.500-g sample of \( _{93}^{237}\mathrm{~Np}\) is # RESULT after calculation of steps 1-5(in millicuries) # . The balanced equation for the decay of \( _{93}^{237}\mathrm{Np}\) is \( _{93}^{237}\mathrm{Np} \rightarrow _{91}^{233}\mathrm{Pa} + _{2}^{4}\mathrm{He} \) .

Step by step solution

01

Determine the Number of Atoms

Use Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol) to find the number of atoms in the sample. Since the atomic mass of neptunium-237 is 237 g/mol, the calculation is:\n\nNumber of Atoms = Mass of Sample x (Avogadro's Number / Atomic Mass) = \(0.500 \, \mathrm{g} \times (6.022 \times 10^{23} \, \mathrm{atoms/mol} / 237 \, \mathrm{g/mol})\)
02

Calculate the Decay Constant

Apply the half-life to find the decay constant (\( \lambda \)), using the formula: \n\n\( \lambda = \ln(2) / \mathrm{Half-life} = \ln(2) / 2.20 \times 10^{6}\) \, years
03

Calculate the Radioactivity

Radioactivity (R) is defined as the rate at which an isotope decays, which can be found by multiplying the decay constant by the number of atoms: \n\nR = \( \lambda \times \mathrm{Number of Atoms}\)
04

Convert the Activity to Curies

We have the radioactivity in decays per year from Step 3. But activity is often reported in curies, where 1 curie (Ci) = \(3.70 \times 10^{10}\) decays per second. To make this conversion, we should convert years to seconds and convert the radioactivity to curies using the conversion factor.
05

Convert the Activity to Millicuries

To convert the activity to millicuries, multiply the activity in curies by 1000, because 1 curie = 1000 millicuries.
06

Write the Nuclear Equation for Alpha decay of Neptunium-237

In alpha decay, the nucleus emits an alpha particle, which has 2 protons and 2 neutrons and is essentially a helium nucleus. So the original atom loses 2 protons and 2 neutrons, becoming a different element. For neptunium-237, this atomic and mass number transition result in the creation of an atom of protactinium-233 and an alpha particle (helium nucleus) - as such: \n\n \( _{93}^{237}\mathrm{Np} \rightarrow _{91}^{233}\mathrm{Pa} + _{2}^{4}\mathrm{He} \)

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Most popular questions from this chapter

Calculate the energy released (in joules) from the following fusion reaction: $${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}$$ The atomic masses are \({ }_{1}^{2} \mathrm{H}=2.0140 \mathrm{amu},{ }_{1}^{3} \mathrm{H}=3.01603\) \(\mathrm{amu},{ }_{2}^{4} \mathrm{He}=4.00260 \mathrm{amu},{ }_{0}^{1} \mathrm{n}=1.008665 \mathrm{amu}\).

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Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear decay. As an example, consider the \(\beta\) decay of \({ }^{90} \mathrm{Sr}\) \((89.907738 \mathrm{amu})\) $${ }_{38}^{90} \mathrm{Sr} \longrightarrow{ }_{39}^{90} \mathrm{Y}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=28.1 \mathrm{yr}$$ The \({ }^{90} \mathrm{Y}\) (89.907152 amu) further decays as follows: $${ }_{39}^{90} \mathrm{Y} \longrightarrow{ }_{40}^{90} \mathrm{Zr}+{ }_{-1}^{0} \beta \quad t_{\frac{1}{2}}=64 \mathrm{~h}$$ Zirconium-90 (89.904703 amu) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the above two decays. (The mass of the electron is \(5.4857 \times 10^{-4}\) amu. \()\) (b) Starting with one mole of \({ }^{90} \mathrm{Sr}\), calculate the number of moles of \({ }^{90} \mathrm{Sr}\) that will decay in a year. (c) Calculate the amount of heat released (in kilojoules) corresponding to the number of moles of \({ }^{90} \mathrm{Sr}\) decayed to \({ }^{90} \mathrm{Zr}\) in \((\mathrm{b})\)
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