It should be no surprise that the Schrödinger result exactly matches the Bohr result. Recall that Bohr was able to explain the spectrum of the H atom, exactly matching the best experimental data available at the time.

Explanation of the spectrum of the H atom

Explanation of the spectrum of the H atom Energy Level Diagram for the H atom

Explanation of the spectrum of the H atom Energy Level Diagram for the H atom Consider two energy levels for the H atom, the upper one characterized by the quantum number nupper (abbreviated nu) and a lower level characterized by the quantum number nlower (abbreviated nl).

Two situations: (1) absorption and (2) emission. nu nl absorption (photon absorbed)

nu nl emission (photon emitted)

In both the absorption and emission processes Recall so that

In both the absorption and emission processes Recall so that k From the formula En = - we can n2 write :

and

and Since we have

and Since we have

and Since we have

and Since we have

That is

That is Set

That is Set So that

That is Set So that This is called the Rydberg formula and RH is called the Rydberg constant.

Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states.

Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states. RH = 109678 cm-1 and k = 2.17869 x 10-18J

Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states. RH = 109678 cm-1 and k = 2.17869 x 10-18J Approach 1

Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states. RH = 109678 cm-1 and k = 2.17869 x 10-18J Approach 1

Problem Example: What is the wavelength and frequency of photons emitted during a transition from nu = 5 state to the nl =2 state in the H atom? Also calculate the energy difference between these two states. RH = 109678 cm-1 and k = 2.17869 x 10-18J Approach 1

= 23032.4 cm-1

= 23032.4 cm-1 Hence = 4.34171 x10-5 cm = 434.171 nm

= 23032.4 cm-1 Hence = 4.34171 x10-5 cm = 434.171 nm Now use

= 4.57528 x 10-19J

= 4.57528 x 10-19J Approach 2 Use k En = - n2

So that k k E5 = - E2 = - 52 22

So that k k E5 = - E2 = - 52 22 = -8.71476 x 10-20 J = -5.44673 x 10-19 J

So that k k E5 = - E2 = - 52 22 = -8.71476 x 10-20 J = -5.44673 x 10-19 J Therefore = E5 - E2 = 4.57525 x 10-19 J

So that k k E5 = - E2 = - 52 22 = -8.71476 x 10-20 J = -5.44673 x 10-19 J Therefore = E5 - E2 = 4.57525 x 10-19 J Now so

So that k k E5 = - E2 = - 52 22 = -8.71476 x 10-20 J = -5.44673 x 10-19 J Therefore = E5 - E2 = 4.57525 x 10-19 J Now so = 4.34174 x 10-7 m

Definition: The ionization energy is the energy required to remove an electron from one mole of a substance in its ground state in the gas phase.

Definition: The ionization energy is the energy required to remove an electron from one mole of a substance in its ground state in the gas phase. For example, for substance X, X(g) X(g)+ + e-

Problem: Calculate the ionization energy for the hydrogen atom. (k = 2.17869 x 10-18J and NA = 6.02214 x 1023 mol-1).

Problem: Calculate the ionization energy for the hydrogen atom. (k = 2.17869 x 10-18J and NA = 6.02214 x 1023 mol-1). The process is: H(g) H+(g) + e-

Problem: Calculate the ionization energy for the hydrogen atom. (k = 2.17869 x 10-18J and NA = 6.02214 x 1023 mol-1). The process is: H(g) H+(g) + e- The quantum numbers are nl = 1 and nu = (because the electron is completely removed from the atom).

Therefore = - E1 = 0 - (-k)

Therefore = - E1 = 0 - (-k) = k = 2.17869 x 10-18J

This corresponds to the energy to remove an electron from one atom. Therefore = - E1 = 0 - (-k) = k = 2.17869 x 10-18J This corresponds to the energy to remove an electron from one atom.

Therefore = - E1 = 0 - (-k) = k = 2.17869 x 10-18J This corresponds to the energy to remove an electron from one atom. To get the ionization potential, we need the energy expended for 1 mole of atoms. Hence

Therefore = - E1 = 0 - (-k) = k = 2.17869 x 10-18J This corresponds to the energy to remove an electron from one atom. To get the ionization potential, we need the energy expended for 1 mole of atoms. Hence Ionization energy = 2.17869 x 10-18 J (6.02214 x 1023 mol-1) = 1.31204 x 106 J mol-1

Definition: The electron affinity is the energy required to add an electron to one mole of a substance in its ground state in the gas phase.

Definition: The electron affinity is the energy required to add an electron to one mole of a substance in its ground state in the gas phase. For example, for substance Y, Y(g) + e- Y(g)-

Exercise: Calculate the electron affinity for the hydrogen positive ion.

Exercise: Calculate the electron affinity for the hydrogen positive ion. (k = 2.17869 x 10-18J and NA = 6.02214 x 1023 mol-1).

Exercise: Calculate the electron affinity for the hydrogen positive ion. (k = 2.17869 x 10-18J and NA = 6.02214 x 1023 mol-1). (By convention the electron affinity is the difference between the energy of the product formed and the energy of the starting species, in this case E(H) – E(H+) ).

How does the Schrödinger eq. do for more complex systems, e. g How does the Schrödinger eq. do for more complex systems, e.g. He atom, CO molecule, etcetera?

How does the Schrödinger eq. do for more complex systems, e. g How does the Schrödinger eq. do for more complex systems, e.g. He atom, CO molecule, etcetera? It turns out that the Schrödinger equation is too complex to solve exactly for systems with more than one electron. However, approximate solutions have been obtained which match up with experimental results.

How does the Schrödinger eq. do for more complex systems, e. g How does the Schrödinger eq. do for more complex systems, e.g. He atom, CO molecule, etcetera? It turns out that the Schrödinger equation is too complex to solve exactly for systems with more than one electron. However, approximate solutions have been obtained which match up with experimental results. For example, some of the energy levels of the He atom have been determined to around 10 significant figures! The ground state energy level for He is known to over 40 significant figures!

It is generally accepted that the Schrödinger equation provides a very good description of the ground and excited states of atomic and molecular systems.

It is generally accepted that the Schrödinger equation provides a very good description of the ground and excited states of atomic and molecular systems. The work of Schrödinger was one of the landmark breakthroughs in all of science.

The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms.

The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms. It is assumed the electronic behavior in many-electron atoms is not too different from that in the H atom.

The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms. It is assumed the electronic behavior in many-electron atoms is not too different from that in the H atom. So the results from the H atom can be used as a first approximation for describing the behavior of the electrons in more complex atoms.

The fact that the H atom can be solved exactly has turned out to be important for studying more complex atoms. It is assumed the electronic behavior in many-electron atoms is not too different from that in the H atom. So the results from the H atom can be used as a first approximation for describing the behavior of the electrons in more complex atoms. The justification for this approach is that it works!

Some final thoughts on the Bohr model

Some final thoughts on the Bohr model Obviously the result obtained by Bohr was very useful to Schrödinger.

Some final thoughts on the Bohr model Obviously the result obtained by Bohr was very useful to Schrödinger. Connection with Heisenberg Uncertainty Principle.

Some final thoughts on the Bohr model Obviously the result obtained by Bohr was very useful to Schrödinger. Connection with Heisenberg Uncertainty Principle. According to Bohr, an electron is always circling around the nucleus in a well specified orbit.