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Chapter 21: Problem 63

A portion of the Sun's energy comes from the reaction $$ 4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{1}^{0} \mathrm{e} $$ which requires a temperature of \(10^{6}\) to \(10^{7} \mathrm{~K}\). Use the mass of the helium-4 nucleus given in Table 21.7 to determine how much energy is released per mol of hydrogen atoms.

Short Answer

Expert verified
The energy released per mole of hydrogen atoms in the given nuclear reaction is approximately \(6.44 \times 10^{11} \, \mathrm{J/mol}\).

Step by step solution

01

Reactants and Products Masses

From the given nuclear reaction: $$4{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{1}^{0}\mathrm{e}$$ We have the following reactants and products: Reactants: 1. 4 Hydrogen atoms (\({ }_{1}^{1} \mathrm{H}\)) Products: 1. 1 Helium-4 nucleus (\({ }_{2}^{4} \mathrm{He}\)) 2. 2 Positrons (\({ }_{1}^{0}\mathrm{e}\)) Step 2: Calculate the mass difference
02

Mass Difference Calculation

First, let's find the total mass of reactants and products. Total mass of reactants: 4 Hydrogen atoms: \(4 \times 1.007825 \, \mathrm{amu} = 4.0313 \, \mathrm{amu}\) Total mass of products: 1 Helium-4 nucleus: \(4.001506 \, \mathrm{amu}\) 2 Positrons: \(2 \times 0.000549 \, \mathrm{amu} = 0.001098 \, \mathrm{amu}\) Total mass of products: \(4.001506+0.001098 = 4.002604 \, \mathrm{amu}\) Mass difference (Reactants - Products) = \(4.0313 - 4.002604 = 0.028696 \, \mathrm{amu}\) Step 3: Convert mass difference to energy
03

Mass Difference to Energy Conversion

We will use Einstein's mass-energy equivalence formula to convert the mass difference to energy. $$E = mc^2$$ Since we have the mass difference in atomic mass units (amu), we first need to convert it to kilograms. Mass difference in kilograms: \(0.028696 \, \mathrm{amu} \times \frac{1.660539 \times 10^{-27} \, \mathrm{kg}}{1 \, \mathrm{amu}} = 4.765 \times 10^{-29} \, \mathrm{kg}\) Now, we can calculate the energy in joules: $$E = (4.765 \times 10^{-29}\, \mathrm{kg}) \times (3 \times 10^8 \, \mathrm{m/s})^2 = 4.289 \times 10^{-12} \, \mathrm{J}\) Step 4: Convert energy to per mole of hydrogen atoms
04

Energy per Mole of Hydrogen

Since the reaction involves 4 hydrogen atoms, we need to find the energy released for 1 mole of hydrogen atoms. Energy per mole of hydrogen atoms: \(\frac{4.289 \times 10^{-12} \, \mathrm{J}}{4 \, \mathrm{hydrogen \, atoms}}\times 6.022\times10^{23}\,\mathrm{atoms/mol} = 6.44 \times 10^{11} \, \mathrm{J/mol}\) The energy released per mole of hydrogen atoms in this reaction is approximately \(6.44 \times 10^{11} \, \mathrm{J/mol}\).

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Most popular questions from this chapter

The Sun radiates energy into space at the rate of $3.9 \times 10^{26} \mathrm{~J} / \mathrm{s} .$ (a) Calculate the rate of mass loss from the Sun in kg/s. (b) How does this mass loss arise? (c) It is estimated that the Sun contains \(9 \times 10^{56}\) free protons. How many protons per second are consumed in nuclear reactions in the Sun?

The thermite reaction, $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(s)+$ $\mathrm{Al}_{2} \mathrm{O}_{3}(s), \Delta H^{\circ}=-851.5 \mathrm{~kJ} / \mathrm{mol}$, is one of the most exothermic reactions known. Because the heat released is sufficient to melt the iron product, the reaction is used to weld metal under the ocean. How much heat is released per mole of $\mathrm{Al}_{2} \mathrm{O}_{3}$ produced? How does this amount of thermal energy compare with the energy released when 2 mol of protons and 2 mol of neutrons combine to form 1 mol of alpha particles?

How much energy must be supplied to break a single \({ }^{21}\) Ne nucleus into separated protons and neutrons if the nucleus has a mass of $20.98846 \mathrm{u}\( ? What is the nuclear binding energy for \)1 \mathrm{~mol}\( of \){ }^{21} \mathrm{Ne}$ ?

Charcoal samples from Stonehenge in England were burned in \(\mathrm{O}_{2},\) and the resultant \(\mathrm{CO}_{2}\) gas bubbled into a solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\) (limewater), resulting in the precipitation of \(\mathrm{CaCO}_{3}\). The \(\mathrm{CaCO}_{3}\) was removed by filtration and dried. A 788 -mg sample of the \(\mathrm{CaCO}_{3}\) had a radioactivity of $1.5 \times 10^{-2}$ Bq due to carbon-14. By comparison, living organisms undergo 15.3 disintegrations per minute per gram of carbon. Using the half-life of carbon-14, 5700 yr, calculate the age of the charcoal sample.

What is the function of the control rods in a nuclear reactor? What substances are used to construct control rods? Why are these substances chosen?
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