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ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Quantum Mechanics > 10. Multi-electron Atoms > The Bohr Atom

As we know it the atom is made up of neutrons, protons, and electrons. While the neutrons (0 charge), and the protons (positive charge) make up the nucleus of the atom, the electron (negative charge) circle around the nucleus much like the orbiting of planets around the sun. According to old theories it was known that electrons are in constant motion around the nucleus while emitting energy. According to this idea as the electrons lost energy they would get closer and closer to the nucleus and eventually crash into it. This problem was solved when Niels Bohr created a model of a hydrogen atom.

- Electrons move around the nucleus of an atom in circular motion.
- Electrons have a set number of orbitals, ring-like paths around the nucleus, they can travel in called
*stationary states*. If the electron stays in one orbital, the energy of the electron remains constant. The first orbital is n=1, the second orbital n=2 and so on. (These values of n are called quantum numbers.) - Electrons can only move from one orbital to another allowed orbital at one time. If an electron drops down from n=2 to n=1 energy is emitted, and if the electron moves up from n=1 to n=2 energy is absorbed. The amount of energy that is either absorbed or emitted is called quanta.

In his model of the atom Bohr used Planck's quantum hypothesis, and of course his knowledge from prior findings. Bohr uses Ryberg's formula for explaining how the electrons emit light as they move from one **orbital** to another. The point Bohr was trying to get across is that energy is not continuous in an atom. We can say that an atom that is in the lowest energy level is in the **ground state,** and when it moves to a higher level it is in an **excited state**. The energy of a photon, lost or gained, is calculated using Planck's equation: (h is Planck's constant, 6.62607 x 10^{-34 }J s / cycles, and v stands for frequency in cycles/s)

\[ \text{Energy Difference} (\Delta E) = h\nu \]

Bohr's model allows us to calculate the radii of the orbits that are allowed for an electron to travel. It also allows us to calculate electron velocities and energy in these orbits. With the equation below we can calculate allowed energy levels for a Hydrogen atom. (-R_{H} is a constant value of 2.179 x 10^{-18 }Joules)

\[ E_n=-R_H \dfrac{1}{n^2} \]

When an electron drops from a higher level for example from n=2 to a lower energy level, n=1, it emits a specific wavelength distinct to that element. To calculate the difference between the two energy levels, the equation below is used. Subscript i stands for initial, (initial energy= E_{i} and initial quantum number= n_{i}) while subscript f stands for final (final energy= E_{f} and final quantum number= n_{f}).

\[ \text{Energy Difference (} \Delta E \text {)} = E_f-E_i = -R_H \left ( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right ) = R_H \left ( \dfrac{1}{n_i^2} - \dfrac{1}{n_f^2}\right )\]

The Bohr model also helps us understand how cations are made. This is through the concept of Ionization Energy, the energy to remove an electron from its ground state. After the electron is free the atom is ionized. E_{i }stands for the ionization energy of the hydrogen atom. The equation below however works for any species that have one electron like hydrogen (for example, He^{+ }and Li^{2}^{+}) where Z is the atomic number of the hydrogen-like atom:

\[ E_n=-Z^2R_H \dfrac{1}{n^2} \]

If we want to find the energy associated with a change in quantum number of a hydrogen-like atom, we can combine this equation with the previous one to form

\[ \Delta E=-Z^2R_H \left ( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right ) \]

If the result of this equation is negative, meaning that the electron is moving down in energy levels (quantum number n), the result is that radiation is emitted. If the result is positive, and the electron is moving up in energy levels (quantum number n), radiation was absorbed.

Although Bohr’s discovery was a huge accomplishment, this model of the atom is not without its flaws. Because it is only designed for the hydrogen atom and hydrogen-like ions, it cannot explain the electron emission of atoms or ions that contain more than one electron.

As stated before, we all now know that the atom is made up of neutrons, protons, and electrons. However, how this knowledge was found and how the Bohr Atom was created cannot be understood without knowledge of the history of the discoveries that occurred before the creation of the Bohr Atom. Although not all of these discoveries are necessary to know, there are a few significant discoveries that are important to know.

In 1808, John Dalton published a book named *A New System of Chemical Philosophy*. In the book, he shared his conclusions on the atom, which became known as Dalton's Atomic Theory. The assumptions of the theory were:

- All matter, or more specifically elements, are made up of tiny, indivisible particles called atoms. And, these atoms, during a chemical reaction, cannot be destroyed nor created.
- The atoms of a specific element are identical (including their mass).
- Elements can be identified based on their individual masses. This is because the masses and properties of the atoms of a specific element are different from the masses and properties of the atoms of a different element.
- When the atoms of elements react, they must combine in simple whole number ratios.

In 1897, J.J. Thompson helped establish electrons as being a negatively charged component of atoms using cathode rays. However, how exactly the electrons fit into the atom was unknown at that time. As seen in the figure right above, Thompson theorized in 1904, that the atom was a positively charged "nebulous cloud" which contained evenly dispersed negatively charged electrons. This became known as the Plum Pudding Model due to its similarity to the popular dessert (with fruit being dispersed throughout gelatin).

In 1887, the Photoelectric Effect was observed by Heinrich Hertz, and then later improved upon by Albert Einstein in 1905. The theory stated that under certain conditions, light, when shined upon a metal, would cause the metal to emit a stream of particles. The particles were found to be electrons, and the number of electrons emitted was found to be based on the frequency of the light. Einstein suggested that each particle of light could be classificed as a photon.

By 1911, Ernest Rutherford and his two assistants Hans Geiger and Ernest Marsden had concluded that the **Plum Pudding Model** was incorrect based on their Gold Foil Experiment. The results of the experiment led to three conclusions about the atom:

- The atom has a very compact nucleus, which is a region in the center of the nucleus which holds a majority of the mass and positive charge of the atom.
- The positive charge of each atom is different from the charge of a different atom. And that the number of electrons inside the atom is equal to the number of protons inside the atom (leading to a neutrally charged atom).
- Negatively charged particles (electrons) orbit the nucleus at any distances (much like planets orbiting the sun).

However, based on electrodynamics, the Rutherford Model was found to be unstable. Seeing the problems with the Rutherford Model, Bohr decided to formulate an atom that would correct the instability problems of the Rutherford Model.

- Petrucci, Ralph H., William S. Harwood, Geoffery F. Herring, and Jeffry D. Madura.
__General Chemistry__. 9th ed. New Jersey: Pearsin Prentice Hall, 2007.

1. What is the change of energy of He^{+} when it moves from energy level n=1 to energy level n=5? Is radiation emitted or absorbed?

First things first: Let's write all of the information we are given and the information we are trying to find.

Information given: n_{i}=1 n_{f}=5

Information that we are trying to find: Change in energy (∆E)

He^{+} is a hydrogen-like atom, since it has the same amount of electrons as hydrogen. However, it has a different number of protons than hydrogen, which gives it a different atomic number, so we need to include the atomic number Z in the equation for the change in energy.

∆E= -Z^{2}R_{H}(1/n_{f}^{2}- 1/n_{i}^{2}) where constant R_{H} = 2.179 X 10^{-18}

∆E = - (2^{2})(2.179 X 10^{-18} J)(1/5^{2} - 1/1^{2})

∆E = (-8.716 X 10^{-18} J)(0.04 - 1.00)

*∆E = 8.367 X 10*^{-18} *J*

since the value of energy change positive, * radiation was absorbed *in this change of energy level

**2. When the frequency associated with an energy change of hydrogen is 6.166 X 10 ^{14} cycles/s and the final energy level is 4, what is the initial energy level?**

**Solution**

First, let's write all of the information we are given along with the information that we are trying to find:

Given information: v = 6.166 X 10^{14} cycles/s n_{f} = 4

Information we want to find: n_{i}

Since this question is referring to a hydrogen atom, we can use the equations

\[\Delta{E}= h\nu\]

and

\[\Delta{E} = - R_H \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)\]

and equate them to each other, resulting in:

\[h\nu = - R_H \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)\]

where constant h = 6.626 X 10^{-34} J s/cycle and constant R_{H}= 2.179 X 10^{-18} J.

So the only variable in this new equation that is not given to us is ni. Let's plug in our known values now:

(6.626 X 10^{-34} J s/cycle) (6.167 X 10^{14} cycles/s)=(-2.179 X 10^{-18} J)[1/(4^{2})- 1/(n_{i}^{2})]

Remember to pay attention to the units in every equation. In these next few steps, we are going to clean up the calculations, and many of these units will cancel out.

(4.086 X 10^{-19} J)/ (-2.179 X 10^{-18} J) = 0.0625 - 1/(n_{i}^{2})

-.1875 - .0625 = -1/(n_{i}^{2})

n_{i}^{2} = 1/.25

n_{i}^{2} = 4

*n*_{i}* = 2*

**3. If there is a frequency of 1.867 X 10 ^{12} s^{-1} involved with a change in energy level, what must the change in energy of this photon be?**

**Solution**

We are given v = 1.867 X 10^{12} s^{-1}

and we are trying to find ∆E.

We know that constant h = 6.626 X 10^{-34} J s/photon, so we can plug these values into the equation:

∆E = hv

∆E = (6.626 X 10^{-34} J s/photon)(1.867 X 10^{12} s^{-1})

__∆E = 12.371 X 10__^{-22} __J/photon__

**4. If possible, what is a likely energy level for a hydrogen atom with E _{n}= -6.053 X 10^{-20} J?**

**Solution**

We are given E_{n} = -6.053 X 10^{-20} J and we are trying to find the value of n.

Since this is question is about a hydrogen atom, we can use the equation:

E_{n}= -R_{H}/n^{2} with constant R_{H}= 2.179 X 10^{-18} J

Since we need to find the value of n to decide if the energy level is likely, we can rearrange this equation to show

n^{2} = -R_{H}/ E_{n}

n^{2} = (-2.179 X 10^{-18} J)/(-6.053 X 10^{-20} J)

n^{2 }= 35.999

n = 6

__Because the value of n is an integer, 6 is the likely energy level of this photon.__

**5. What is the energy change of a hydrogen atom when it moves from energy level n=4 to energy level n=1? Is radiation emitted or absorbed? What is the wavelength for this change?**

**Solution**

The values we are given: n_{i} = 4 n_{f} = 1

What we are trying to find: ∆ E and wavelength

Since this question asks about energy changes of a hydrogen atom, we can use this equation to find the energy change:

∆ E = - R_{H}( 1/n_{f}^{2} - 1/n_{i}^{2})

And once we know the value of ∆ E, we can combine the two equations

∆ E = hv and v = c /wavelength to get the equation ∆ E = hc/wavelength

where constant h = 6.626 X 10^{-34} J s/photon and constant c = 2.9979 X 10^{8} m/s

So, first let's find the value of ∆ E:

∆ E = (-2.179 X 10^{-18} J)(1/1^{2} - 1/4^{2})

__∆ E = - 2.043 X 10__^{-18} __J__

Since this photon is moving down in energy levels and the value of ∆ E is negative...

__Radiation was emitted.__

Now, let's find the wavelength

since ∆ E = hc/wavelength, wavelength = hc/∆ E, so

wavelength = (6.626 X 10^{-34} J s/photon)(2.9979 X 10^{8} m/s)/ (-2.043 X 10^{-18} J/photon)

__wavelength = - 9.723 X 10__^{-8}__ m, which is -97.23 nm__

**6. What is the wavelength of the photon required to completely ionize a hydrogen atom initially in the energy level: n=5?**

The values we are given: n_{i} = 5 n_{f} = ∞

We know that:

∆ E = - R_{H}( 1/n_{f}^{2} - 1/n_{i}^{2})

∆ E = hv and v = c /wavelength to get the equation ∆ E = hc/wavelength

- R_{H = }-2.179 X 10^{-18} J

∆ E = (-2.179 X 10^{-18} J) ( (1/∞^{2}) - (1/5^{2}) )

∆ E = 8.716 X 10^{-19} J

∆ E = hc/wavelength

8.716 X 10^{-19} J = (6.626 X 10^{-34} Js) (2.9979 X 10^{8} m/s) / wavelength

__Wavelength = 2.279 X 10 ^{-7} m = 227.9 nm__

- Forogh Rahim (UCD), Antoinette Mursa (UCD)

Last Modified

09:00, 16 Nov 2014

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