
The Bohr model for the hydrogen atom allowed electrons in atoms only to occupy particular, discrete orbits corresponding to quantized energy levels. Bohr could not explain why electrons favored these particular orbits; de Broglie proposed that the electron behaves like a wave at the atomic scale, so electron orbits correspond to circular standing waves or resonances. Quantized energy levels provide a direct explanation for absorption and emission of photons by atoms. Emission lines are produced by an atom when an electron makes a transition from a higher to a lower energy level. Absorption lines occur when the electron in an atom absorbs a photon with a specific energy, moving it from a lower to a higher energy level. An absorptionline spectrum appears when atoms absorb particular wavelengths of light from the spectrum of a background light source.

phosphorescence, orbit, Bohr model, quantum number, quantized, energy levels, ground state, electron volt (eV), excited state, absorption, emission, scattering, spectral line, line spectrum, continuum spectrum, spectrograph, Pauli exclusion principle

Review problems and questions 


By using the Bohr model of the atom, should the size of the helium atom be larger or smaller than that of the hydrogen atom? Why?

The Bohr model of the atom really only applies to hydrogenic atoms, i.e., those with only one electron, so it could apply to singly ionized helium. The basic concept of the Bohr model is that the orbit of the electron is maintained by the electric force of attraction between the positively charged nucleus and the negatively charged electron. If the positive charge of the nucleus doubles, then the electric force also doubles, pulling the electron closer to the nucleus. This results in a smaller radius for the helium atom.


One hydrogen atom has an electron in its n = 1 energy level, while another atom has an electron in its n = 2 level. Which one requires more energy to move the electron to the n = 3 level? To the n = ∞ level? Why?

The difference between the n = 1 and n = 3 energy levels is greater than the difference between the n = 2 and n = 3 levels. More energy is therefore required to cause the electron in the n = 1 energy level to transition to the n = 3 level. The same applies for transitions to any higher n level, including n = ∞.


What property of atoms produces an emissionline spectrum?

Emission spectra are the result of quantized electron energy levels in the atom. The transition of an electron from a higher energy level to a lower energy level causes the emission of a photon with a corresponding energy (and frequency). The many possible transitions of this kind can produce a spectrum of emission lines.


The hydrogen atom produces three distinct sets of spectral lines known as the Lyman, Balmer, and Paschen series.
 Describe the differences in the light emitted in these three series.
 Explain how these differences are related to the energy level transitions of the atom.

Answer:  The Lyman series spectral lines consist of highenergy ultraviolet light, the Balmer series lines lie in the visible range, and the Paschen series consists of lowenergy infrared light.
 Photons that form the highenergy Lyman series lines are emitted during transitions down to the lowest energy level, n = 1. The Balmer series involves transitions to the n = 2 level. The Paschen series is produced by transitions to the n = 3 level.


The electron in an atom changes energy levels, and in the process it emits a photon at a wavelength of 656.3 nm.
 Is the electron energy level higher before or after it emits the photon?
 In joules, what is the difference in energy between the two energy levels?
 What is this energy difference in electron volts?

Answer:  The level is higher before.
 3.031×10^{−19} J
 1.892 eV
Solution: If the atom emits a photon, it loses energy equal to the energy of the emitted photon. The energy level before emission is therefore higher than the energy level afterward.
 Asked: the difference in energy between the energy levels the electron moved
Given: wavelength of the emitted photon, λ = 656.3 nm Relationships: c = fλ; E = hf; speed of light c = 3×10^{8} m/s; Planck's constant h = 6.63×10^{−34} J s Solve: $$c=f\lambda $$Rearrange to solve for f:$$f=\frac{c}{\lambda}$$Substitute f into E = hf:$$E=h\frac{c}{\lambda}$$Solve for E:$$\begin{array}{lll}E\hfill & =\hfill & h\frac{c}{\lambda}\hfill \\ \hfill & =\hfill & (6.63\times {10}^{34}\text{Js})\frac{3\times {10}^{8}\text{m/s}}{656.3\times {10}^{9}\text{m}}\hfill \\ \hfill & =\hfill & 3.031\times {10}^{19}\text{J}\hfill \end{array}$$Answer: The difference in energy is 3.031×10^{−19} J  Asked: energy difference in electron volts
Given: change in energy, ΔE = 3.031×10^{−19} J Relationships: 1 eV = 1.602×10^{−19} J Solve: $$\left(3.031\times {10}^{19}\text{J}\right)\times \left(\frac{1\text{eV}}{1.602\times {10}^{19}\text{J}}\right)=1.892\text{eV}$$Answer: The difference in energy is 1.892 eV.


An atom absorbs a photon of light that has a frequency of 10^{14} Hz.
 What is the energy of this photon of light?
 What is the energy difference between the energy level of the atom before it absorbs the photon and the energy level after it absorbs the photon?
 What part of the electromagnetic spectrum does this photon correspond to?

Answer:  6.63×10^{−20} J
 +6.63×10^{−20} J
 infrared radiation
Solution: Asked: energy of this photon
Given: frequency f = 10^{14} Hz Relationships: E = h f, Planck’s constant h = 6.63×10^{34} J s Solve:$$E=hf=(6.63\times {10}^{34}\text{Js})\times {10}^{14}\text{Hz}=6.63\times {10}^{20}\text{J}$$Answer: The energy of this photon is 6.63×10^{−20} J  The difference in energy levels is the same as the energy of the absorbed photon of light, i.e., 6.63×10^{−20} J. (The final energy for the atom is higher than the initial energy, because the atom absorbed energy from the photon.)
 Taking the frequency of the photon f = 10^{14} Hz, and comparing with the figure of the electromagnetic spectrum, tells us that the photon is infrared light.

 Two students were each trying to use the same type of spectrograph. One of them read in the instruction manual never to point the spectrograph at the Sun. The other student found instructions on the Internet for a spectrograph that said you could point it at the Sun. Which instructions should they follow? Why?

The students should not point the spectrograph at the Sun. The instructions the second student found might apply to a different type of spectrograph from the kind that they were using. Always follow the instruction manual that comes with a scientific instrument, not the instructions found possibly at random on the Internet!

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