The sun is surrounded by a white circle of gaseous material called the corona, which becomes visible during a total eclipse of the sun. The temperature of the corona is in the millions of degrees Celsius, which is high enough to break up molecules and remove some or all of the electrons from atoms. One way astronomers have been able to estimate the temperature of the corona is by studying the emission lines of ions of certain elements. For example, the emission spectrum of \(\mathrm{Fe}^{14+}\) ions has been recorded and analyzed. Knowing that it takes \(3.5 \times 10^{4} \mathrm{~kJ} / \mathrm{mol}\) to convert \(\mathrm{Fe}^{13+}\) to \(\mathrm{Fe}^{14+},\) estimate the temperature of the sun's corona.

Step by step solution

01

Convert energy unit from kJ/mol to J/ion

The energy given in the problem, \(3.5 \times 10^{4}~ \mathrm{kJ} / \mathrm{mol}\), has to be converted to J/ion, the unit used in the Boltzmann equation formulas, to be useful. The molar mass of iron (Fe) is 55gmol and Avogadro's number is \(6.022\times10^{23}\) ions/mole. The conversion is given thus: \[3.5 \times 10^{4}\, \mathrm{kJ/mol} = 3.5 \times 10^{7}\, \mathrm{J/mol} = \frac{3.5 \times 10^{7}\, \mathrm{J/mol}}{6.022\times10^{23}\, \mathrm{ions/mol}} = \frac{5.81 \times 10^{-17}}{\mathrm{ion}} \, J/ion\].

02

Use the Boltzmann equation

Boltzmann's equation can be rearranged to give: \[T=\frac{E}{k_B \times ln(2)}\], where \(E\) is the energy change, \(k_B\) is Boltzmann's constant and \(T\) is the temperature. Here, Boltzmann's constant \(k_B\) is \(1.38 \times 10^{-23}\, J/K\). Substituting the energy change \(E\) from step 1 and Boltzmann's constant \(k_B\) into the equation yields the temperature of the corona.

03

Calculate the temperature of the corona

Substitute the energy change and Boltzmann's constant into the equation from step 2: \[T= \frac{E}{k_B \times ln(2)} = \frac{5.81 \times 10^{-17}\, J/ion}{1.38 \times 10^{-23}\, J/K \times ln(2)}\]. After calculation, we got the temperature of the corona to be approximately \(3.04 \times 10^6 K\) (kelvin), which is roughly equal to the temperature in degrees Celsius.

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