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Chapter 7: Problem 34

In the hydrogen atom, what is the physical significance of the state for which \(n=\infty\) and \(E=0\) ?

Short Answer

Expert verified
The physical significance of a hydrogen atom state with \(n=\infty\) and \(E=0\) is that the electron is no longer bound to the nucleus, having enough energy to escape the atom's influence. In this scenario, the atom becomes an ion, with the electron far from the nucleus and practically not interacting with it anymore. It represents the highest energy level for the atom, right before complete ionization.

Step by step solution

01

Principal Quantum Number and Energy Levels

In the hydrogen atom, the energy levels are dependent on the principal quantum number \(n\), which can take integer values starting from 1. The energy is given by the formula: \[E = \frac{-13.6 \,\text{eV}}{n^2}\] The smaller the quantum number \(n\), the lower the energy level will be, and the electron will be closer to the atomic nucleus. As the quantum number \(n\) increases, the energy level becomes less negative, causing the electron to move farther from the nucleus, and the atom's size expands.
02

State with \(n=\infty\) and \(E=0\)

When the quantum number \(n\) approaches infinity, the energy of the state tends to zero: \[\lim_{n \to \infty} E = \lim_{n \to \infty} \frac{-13.6 \,\text{eV}}{n^2} = 0\] In this scenario, the electron is no longer bounded to the nucleus and has sufficient energy to escape the atom's influence. The electron is far from the nucleus, practically not interacting with it anymore, and the atom is ionized. The hydrogen atom with \(n=\infty\) and \(E=0\) represents the most energetic state 鈥� an atom on the edge of ionization.
03

Physical significance

In summary, the physical significance of a hydrogen state with \(n=\infty\) and \(E=0\) is that the electron is no longer bounded to the nucleus, with enough energy to escape the atom's grasp. The atom becomes an ion, with the electron far from the nucleus and essentially not interacting with it anymore. This is the highest energy level for the atom, right before complete ionization.

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Most popular questions from this chapter

Give the maximum number of electrons in an atom that can have these quantum numbers: a. \(n=0, \ell=0, m_{\ell}=0\) b. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-\frac{1}{2}\) c. \(n=3, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=2\) e. \(n=1, \ell=0, m_{\ell}=0\)

One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for \(\mathrm{Li}, \mathrm{N}, \mathrm{Ni}, \mathrm{Te}, \mathrm{Ba}\), and \(\mathrm{Hg} .\) Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

Draw atomic orbital diagrams representing the ground-state electron configuration for each of the following elements. a. \(\mathrm{Na}\) b. \(\mathrm{Co}\) c. \(\mathrm{Kr}\) How many unpaired electrons are present in each element?

What are the possible values for the quantum numbers \(n, \ell\), and \(m_{\ell} ?\)

From the information below, identify element \(\mathrm{X}\). a. The wavelength of the radio waves sent by an FM station broadcasting at \(97.1 \mathrm{MHz}\) is \(30.0\) million \(\left(3.00 \times 10^{7}\right)\) times greater than the wavelength corresponding to the energy difference between a particular excited state of the hydrogen atom and the ground state. b. Let \(V\) represent the principal quantum number for the valence shell of element \(X\). If an electron in the hydrogen atom falls from shell \(V\) to the inner shell corresponding to the excited state mentioned above in part a, the wavelength of light emitted is the same as the wavelength of an electron moving at a speed of \(570 . \mathrm{m} / \mathrm{s}\) c. The number of unpaired electrons for element \(\mathrm{X}\) in the ground state is the same as the maximum number of electrons in an atom that can have the quantum number designations \(n=2\), \(m_{\ell}=-1\), and \(m_{s}=-\frac{1}{2}\) d. Let \(A\) equal the charge of the stable ion that would form when the undiscovered element 120 forms ionic compounds. This value of \(A\) also represents the angular momentum quantum number for the subshell containing the unpaired electron(s) for element \(\mathrm{X}\).
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