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Chapter 24: Problem 78

How many \(\mathrm{MeV}\) are released per nucleus when \(3.2 \times 10^{-3} \mathrm{~mol}\) of chromium- 49 releases \(8.11 \times 10^{5} \mathrm{~kJ} ?\)

Short Answer

Expert verified
Approximately \(2.63 \times 10^{6} \text{ MeV}\) are released per nucleus.

Step by step solution

01

Calculate the Total Energy Released

Multiply the energy released per mole by the number of moles to find the total energy released.Since the energy released is given as \( 8.11 \times 10^{5} \text{ kJ} \), use this value directly as the total energy.
02

Convert kJ to MeV

Convert the total energy from kilojoules (kJ) to mega electron volts (MeV).Use the conversion factor: \( 1 \text{ kJ} = 6.242 \times 10^{21} \text{ MeV} \).Thus, \( 8.11 \times 10^{5} \text{ kJ} \times 6.242 \times 10^{21} \text{ MeV/kJ} = 5.063 \times 10^{27} \text{ MeV}. \)
03

Calculate Number of Nuclei

Determine the number of nuclei in the given amount of chromium-49 using Avogadro's number.Avogadro's number is \( 6.022 \times 10^{23} \text{ nuclei/mol} \).Thus, the number of nuclei is \( 3.2 \times 10^{-3} \text{ mol} \times 6.022 \times 10^{23} \text{ nuclei/mol} = 1.927 \times 10^{21} \text{ nuclei} \).
04

Calculate Energy Released Per Nucleus

Divide the total energy in MeV by the number of nuclei to find the energy released per nucleus.\( \frac{5.063 \times 10^{27} \text{ MeV}}{1.927 \times 10^{21} \text{ nuclei}} \approx 2.63 \times 10^{6} \text{ MeV/nucleus} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

MeV calculation
When calculating the energy released in nuclear reactions, it's crucial to understand the steps involved in converting kilojoules (kJ) to mega electron volts (MeV). This problem involves the energy released per mole of a substance, which is first given in kJ. To convert this energy into MeV, we must use the conversion factor: 1 kJ = 6.242 x 10^21 MeV. By multiplying the total energy in kJ by this factor, we are able to express the energy in MeV. For example, if 8.11 x 10^5 kJ of energy is released, multiplying this value by 6.242 x 10^21 MeV/kJ gives us approximately 5.063 x 10^27 MeV. This step is essential to solve the problem accurately.
Avogadro’s number
Nuclear chemistry often requires us to calculate the number of atoms or nuclei involved in a reaction. Avogadro's number, which is 6.022 x 10^23 nuclei/mol, helps us achieve this. In the example, we have 3.2 x 10^-3 mol of chromium-49. By multiplying this amount by Avogadro's number, we find the total number of nuclei: 3.2 x 10^-3 mol x 6.022 x 10^23 nuclei/mol = 1.927 x 10^21 nuclei. Knowing the number of nuclei is necessary for calculating the energy released per nucleus.
Energy conversion factors
Understanding and using the correct energy conversion factors is critical in nuclear chemistry. For instance, converting energy from kilojoules to mega electron volts involves using the specific conversion factor: 1 kJ = 6.242 x 10^21 MeV. These conversion factors allow chemists to express energy in the desired units for more straightforward calculations. In practical applications, these conversions enable scientists to compare and measure the energy produced or consumed in nuclear reactions accurately.
Nuclear chemistry
Nuclear chemistry focuses on the chemical processes involving changes in atomic nuclei. It includes studying nuclear reactions, which often release or absorb vast amounts of energy. In this example, chromium-49 undergoes a reaction releasing 8.11 x 10^5 kJ per mole, which translates to significant energy release per nucleus. Such reactions are fundamental to understanding nuclear power, medical treatments using radiation, and various applications requiring precise energy calculations.

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Most popular questions from this chapter

Plutonium- 242 decays to uranium- 238 by emission of an \(\alpha\) particle with an energy of \(4.853 \mathrm{MeV}\). The \({ }^{238} \mathrm{U}\) that forms is unstable and emits a \(\gamma\) ray \((\lambda=0.02757 \mathrm{nm}) .\) (a) Write balanced equations for these reactions. (b) What would be the energy of the \(\alpha\) particle if \({ }^{242} \mathrm{Pu}\) decayed directly to the more stable \({ }^{238} \mathrm{U} ?\)

A sample of \(\mathrm{AgCl}\) emits \(175 \mathrm{nCi} / \mathrm{g} .\) A saturated solution prepared from the solid emits \(1.25 \times 10^{-2} \mathrm{~Bq} / \mathrm{mL}\) due to radioactive\(\mathrm{Ag}^{+}\) ions. What is the molar solubility of \(\mathrm{AgCl} ?\)

Why might it be difficult to use only a nuclide's \(N / Z\) ratio to predict whether it will decay by \(\beta^{+}\) emission or by e \(^{-}\) capture? What other factor is important?

Nuclear disarmament could be accomplished if weapons were not "replenished." The tritium in warheads decays to helium with a half-life of 12.26 yr and must be replaced or the weapon is useless. What fraction of the tritium is lost in 5.50 yr?

Describe the nature and purpose of these components of a nuclear reactor: (a) control rods; (b) moderator; (c) reflector.
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