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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 – 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

Identify the following elements. a. An excited state of this element has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{5} 3 s^{1}\). b. The ground-state electron configuration is \([\mathrm{Ne}] 3 s^{2} 3 p^{4}\). c. An excited state of this element has the electron configuration \([\mathrm{Kr}] 5 s^{2} 4 d^{6} 5 p^{2} 6 s^{1}\) d. The ground-state electron configuration contains three unpaired \(6 p\) electrons.

Without looking at data in the text, sketch a qualitative graph of the third ionization energy versus atomic number for the elements Na through Ar, and explain your graph.

Write the expected electron configurations for each of the following atoms: \(\mathrm{Sc}, \mathrm{Fe}, \mathrm{P}, \mathrm{Cs}\), Eu, \(\mathrm{Pt}, \mathrm{Xe}, \mathrm{Br}\).

Write the expected ground-state electron configuration for each of the following. a. the lightest halogen atom b. the alkali metal with only \(2 p\) and \(3 p\) electrons c. the Group \(3 \mathrm{~A}\) element in the same period as \(\mathrm{Sn}\) d. the nonmetallic elements in Group \(4 \mathrm{~A}\)

Photosynthesis uses \(660-\mathrm{nm}\) light to convert \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) into glucose and \(\mathrm{O}_{2}\). Calculate the frequency of this light.
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