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Chapter 7: Problem 68

Why do the \(3 s, 3 p,\) and \(3 d\) orbitals have the same energy in a hydrogen atom but different energies in a many-electron atom?

Short Answer

Expert verified
In a hydrogen atom, the energy of orbitals only depends on the principal quantum number, hence all orbitals in the third energy level have the same energy. In a multi-electron atom, besides the principal quantum number, the shape of the orbital (determined by the azimuthal quantum number) causes the difference in energy due to electron-electron interactions. This interaction leads to splitting of energies of the \(3s\), \(3p\), and \(3d\) orbitals.

Step by step solution

01

Understanding orbitals in a hydrogen atom

In a hydrogen atom, which only has one electron, the energy of an electron in an orbital depends solely on the principal quantum number \(n\). Thus, the \(3s\), \(3p\), and \(3d\) orbitals, all of which are in the third energy level (\(n=3\)), have the same energy.
02

Understanding orbitals in a multi-electron atom

In a multi-electron atom, the energy of an electron in an orbital depends not only on the principal quantum number \(n\), but also on the azimuthal quantum number \(l\). The \(l\) carries information about the shape of the orbital and electron-electron interactions change based on this shape. This interaction between electrons in different orbitals, also known as electron-electron repulsion, leads to a splitting of energy levels which causes the \(3s\), \(3p\), and \(3d\) orbitals to have different energies.
03

Summary

In hydrogen, the single electron is distant from the nucleus for \(3s\), \(3p\), \(3d\) the same average distance. Thus, they have the same energy which is determined by only the principal quantum number. In a multi-electron atom, electron-electron interactions lead to a difference in energy between orbitals with the same principal quantum number but with different shapes.

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Most popular questions from this chapter

The wave function for the \(2 s\) orbital in the hydrogen atom is $$ \Psi_{2 s}=\frac{1}{\sqrt{2 a_{0}^{3}}}\left(1-\frac{\rho}{2}\right) e^{-\rho / 2} $$ where \(a_{0}\) is the value of the radius of the first Bohr orbit, equal to \(0.529 \mathrm{nm}, \rho\) is \(Z\left(r / a_{0}\right),\) and \(r\) is the distance from the nucleus in meters. Calculate the location of the node of the \(2 s\) wave function from the nucleus.

The ground-state electron configurations listed here are incorrect. Explain what mistakes have been made in each and write the correct electron configurations. Al: \(1 s^{2} 2 s^{2} 2 p^{4} 3 s^{2} 3 p^{3}\) \(\mathrm{B}: 1 s^{2} 2 s^{2} 2 p^{5}\) \(\mathrm{F}: 1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6}\)

Describe the four quantum numbers used to characterize an electron in an atom.

In an electron microscope, electrons are accelerated by passing them through a voltage difference. The kinetic energy thus acquired by the electrons is equal to the voltage times the charge on the electron. Thus, a voltage difference of \(1 \mathrm{~V}\) imparts a kinetic energy of \(1.602 \times 10^{-19} \mathrm{C} \times \mathrm{V}\) or \(1.602 \times\) \(10^{-19} \mathrm{~J}\). Calculate the wavelength associated with electrons accelerated by \(5.00 \times 10^{3} \mathrm{~V}\).

$$ \begin{array}{lccccc} \lambda(\mathrm{nm}) & 405 & 435.8 & 480 & 520 & 577.7 \\ \hline \mathrm{KE}(\mathrm{J}) & 2.360 \times & 2.029 \times & 1.643 \times & 1.417 \times & 1.067 \times \\ & 10^{-19} & 10^{-19} & 10^{-19} & 10^{-19} & 10^{-19} \end{array} $$ A ruby laser produces radiation of wavelength \(633 \mathrm{nm}\) in pulses whose duration is \(1.00 \times 10^{-9} \mathrm{~s}\). (a) If the laser produces \(0.376 \mathrm{~J}\) of energy per pulse, how many photons are produced in each pulse? (b) Calculate the power (in watts) delivered by the laser per pulse. \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s})\)
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