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Chapter 36: Problem 61

(a) Verify Equation 36.8 by considering a single-electron atom with nuclear charge \(\mathrm{Ze}\) instead of \(e\). (b) Calculate the ionization energies for single-electron versions of helium, oxygen, lead, and uranium.

Short Answer

Expert verified
The verification of equation 36.8 with a single-electron atom having nuclear charge \(\mathrm{Ze}\) indeed holds. The ionization energies resultant for single-electron versions of helium, oxygen, lead, and uranium are proportional to the square of their atomic numbers.

Step by step solution

01

Verifying the Equation

Replace \(e\) with \(\mathrm{Ze}\) in the equation 36.8. After substitution check if the equation still holds.
02

Single-electron Ionization Energy Calculation

The general equation for ionization energy is given by \(-13.6 \times \frac{\mathrm{Z}^2}{n^2}\) eV. For one-electron species, \(n = 1\), thus reducing the equation to \(-13.6 \times \mathrm{Z}^2\) eV. Calculate the energies for helium, oxygen, lead, and uranium by substituting their atomic numbers (\(\mathrm{Z}\)) into this equation: helium (\(\mathrm{Z} = 2\)), oxygen (\(\mathrm{Z} = 8\)), lead (\(\mathrm{Z} = 82\)), and uranium (\(\mathrm{Z} = 92\)).
03

Interpret the Result

Compare and analyze the ionization energies calculated for all four elements. They should increase with the increase in atomic number, because the nuclear charge \(\mathrm{Z}\) directly increases the ionization energy.

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