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Chapter 7: Problem 155

Atoms of an element have only two accessible excited states. In an emission experiment, however, three spectral lines were observed. Explain. Write an equation relating the shortest wavelength to the other two wavelengths.

Short Answer

Expert verified
The three spectral lines are observed due to three different electron transitions to the ground state. The shortest wavelength will be produced by the transition from the highest excited state to the ground state. The relationship between the shortest wavelength and the other two wavelengths is given by \(1 / λ_A = 1 / λ_B + 1 / λ_C\).

Step by step solution

01

Understand the Electronic Transitions

There are two energy states that an electron can occupy above the ground state. Essentially, this will give rise to three possible transitions: from the highest energy level to the ground state (Transition A), the next highest energy level to the ground state (Transition B), and between the two excited states to the ground state (Transition C). The transition that involves the greatest energy change will correspond to the shortest wavelength, due to the inverse relationship between energy and wavelength in the equation \(E = h \times c / λ\), where \(E\) is energy, \(h\) is Planck's constant, \(c\) is the speed of light and \(λ\) is the wavelength.
02

Identify the Shortest Wavelength

The greatest energy change (and therefore shortest wavelength) will occur for Transition A, from the highest excited state directly to the ground state.
03

Relate the Wavelengths

The energy of each transition can also be related to the energies of the other transitions. This is due to energy conservation where the energy released in one transition should equal to the sum of the energies released in the other transitions. Therefore, the energy (and thus the frequency and inversely the wavelength) of Transition A can be written as the sum of the energies of Transitions B and C. Using the relationship \(E = h \times c / λ\), we have \(h \times c / λ_A = h \times c / λ_B + h \times c / λ_C\). The Planck's constant and the speed of light can be cancelled from both sides, leaving an equation that relates the wavelengths: \(1 / λ_A = 1 / λ_B + 1 / λ_C\).

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Most popular questions from this chapter

The electron configurations described in this chapter all refer to gaseous atoms in their ground states. An atom may absorb a quantum of energy and promote one of its electrons to a higher-energy orbital. When this happens, we say that the atom is in an excited state. The electron configurations of some excited atoms are given. Identify these atoms and write their ground-state configurations: (a) \(1 s^{1} 2 s^{1}\) (b) \(1 s^{2} 2 s^{2} 2 p^{2} 3 d^{1}\) (c) \(1 s^{2} 2 s^{2} 2 p^{6} 4 s^{1}\) (d) \([\mathrm{Ar}] 4 s^{1} 3 d^{10} 4 p^{4}\) (e) \([\mathrm{Ne}] 3 s^{2} 3 p^{4} 3 d^{1}\)

Explain the statement, Matter and radiation have a "dual nature."

Calculate the frequency (Hz) and wavelength (nm) of the emitted photon when an electron drops from the \(n=4\) to the \(n=2\) level in a hydrogen atom.

Draw orbital diagrams for atoms with the following electron configurations: (a) \(1 s^{2} 2 s^{2} 2 p^{5}\) (b) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}\) (c) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{7}\)

Indicate the number of unpaired electrons present in each of the following atoms: \(\mathrm{B}, \mathrm{Ne}, \mathrm{P}, \mathrm{Sc}, \mathrm{Mn}, \mathrm{Se}\) \(\mathrm{Kr}, \mathrm{Fe}, \mathrm{Cd}, \mathrm{I}, \mathrm{Pb}\)
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