28.3 THE BOHR THEORY OF HYDROGENAt the beginning of the 20th century, scientists were

puzzled by the failure of classical physics to explain the characteristics of spectra

- Why did atoms of a given element emit only certain lines?

- Further, why did the atoms absorb only those wavelengths that they emitted?

In 1913 Bohr provided an explanation of atomic spectra that includes some features of the currently accepted theory. Using the simplest atom, hydrogen, Bohr developed a model of what he thought must be the atom’s structure in an attempt to explain why the atom was stable.

• His model of the hydrogen atom contains some classical features, as well as some revolutionary postulates that could not be justified within the framework of classical physics. The basic assumptions of the Bohr Theory as it applies to the hydrogen atom are as follows:

1- The electron moves in circular orbits about the proton under the influence of the Coulomb force of attraction, as in Figure 28.5. The Coulomb force produces the electron’s centripetal acceleration.

2. Only certain electron orbits are stable. These are orbits in which the hydrogen atom doesn’t

emit energy in the form of electromagnetic radiation. Hence, the total energy of the atom remains constant, and classical mechanics can be used to describe the electron’s motion.

3. Radiation is emitted by the hydrogen atom when the electron “jumps” from a more energetic initial state to a less energetic state. The “jump” can’t be visualized or treated classically.

• In particular, the frequency f of the radiation emitted in the jump is related to the change in the atom’s energy and is independent of the frequency of the electron’s orbital motion. The frequency of the emitted radiation is given by

• where Ei is the energy of the initial state, Ef is the energy of the final state, h is Planck’s constant, and Ei > Ef

• The size of the allowed electron orbits is determined by a condition imposed on the electron’s orbital angular momentum: the allowed orbits are those for which the electron’s orbital angular momentum about the nucleus is an integral multiple of ħ (pronounced “h bar”), where ħ=h/2π :

28.4With these four assumptions, we can calculate the

allowed energies and emission wavelengths of the hydrogen atom. We use the model pictured in Figure 28.5, in which the electron travels in a circular orbit of radius r with an orbital speed v

• The electrical potential energy of the atom is 28.4• where is the Coulomb constant. Assuming the

nucleus is at rest, the total energy E of the atom is the sum of the kinetic and potential energy: 28.5

• We apply Newton’s second law to the electron. We know that the electric force of attraction on the electron, ke e 2/r 2, must equal me ar, which is the centripetal acceleration of the electron. then

• Thus, 28.6

• From this equation, we see that the kinetic energy of the electron is

28.7We can combine this result with Equation 28.5 and

express the energy of the atom as 28.8where the negative value of the energy indicates that

the electron is bound to the proton.An expression for r is obtained by solving Equations 28.4

and 28.6 for v and equating the results: L =mevr = nh/2π =nħ

28.9This equation is based on the assumption that the

electron can exist only in certain allowed orbits determined by the integer n.

• The orbit with the smallest radius, called the Bohr radius, a 0, corresponds to n =1 and has the value 28.10

• A general expression for the radius of any orbit in the hydrogen atom is obtained by substituting Equation 28.10 into Equation 28.9:

28.11

• The first three Bohr orbits for hydrogen are shown in Active Figure 28.6.

• Equation 28.9 may be substituted into Equation 28.8 to give the following expression for the energies of the quantum states:

28.12If we insert numerical values into Equation 28.12,

we obtain 28.13

• The lowest energy state, or ground state, corresponds to n = 1 and has an energy

E1 =-meke2 e 4/2ħ2 = -13.6 eV. The next state,

corresponding to n =2, has an energy E2 = E1/4 = -3.40 eV, and so on. An energy level diagram

showing the energies of these stationary states and the corresponding quantum numbers is given in Active Figure 28.7.

• EXAMPLE 28.1• Goal Calculate the wavelength, frequency, and energy of a

photon emitted during an electron transition in an atom.• Problem The Balmer series for the hydrogen atom corresponds

to electronic transitions that terminate in the state with quantum number n = 2, as shown in Figure 28.8.

• (a) Find the longest-wavelength photon emitted in the Balmer series and determine its frequency and energy. (b) Find the shortest-wavelength photon emitted in the same series.

• Strategy This is a matter of substituting values into Equation 28.17. The frequency can then be obtained from c = fλ and the energy from E= hf. The longest wavelength photon corresponds to the one that is emitted when the electron jumps from the ni = 3 state to the nf = 2 state. The shortest wavelength photon corresponds to the one that is emitted when the electron jumps from ni = ∞ to the state nf = 2.

• Solution• Find the longest wavelength photon emitted in the Balmer

series, and determine its energy.• Substitute into Equation 28.17, with ni = 3 and nf = 2:

• Take the reciprocal and substitute, finding the longest-wavelength :

Now use c = fλ to obtain the frequency:

Calculate the photon’s energy by substituting into Equation 27.5:

• Find the shortest wavelength photon emitted in the Balmer series.

• Substitute into Equation 28.17, with ni = ∞ and nf = 2.

• Take the reciprocal and substitute, finding the wavelength:

• We could also obtain the energy of the photon by using Equation 28.3 in the form hf = E3 - E2, where E2 and E3 are the energy levels of the hydrogen atom

• Bohr’s Correspondence Principle

• we found that Newtonian mechanics cannot be used to describe phenomena that occur at speeds approaching the speed of light.

• Newtonian mechanics is a special case of relativistic mechanics and applies only when v is much smaller than c.

• Similarly, quantum mechanics is in agreement with classical physics when the energy differences between quantized levels are very small. This principle, first set forth by Bohr, is called the correspondence principle.

• For example, consider the hydrogen atom with n > 10000. For such large values of n, the energy differences between adjacent levels approach zero and the levels are nearly continuous, as Equation 28.13 shows. Calculations show that for n > 10000, this frequency is different from that predicted by quantum mechanics by less than 0.015%.

28.4 MODIFICATION OF THE BOHR THEORYThe Bohr Theory of the hydrogen atom was a

tremendous success in certain areas because 1. it explained several features of the hydrogen

spectrum that had previously defied explanation2. It accounted for the Balmer series and other series3. it predicted a value for the Rydberg constant that

is in excellent agreement with the experimental value

4. it gave an expression for the radius of the atom; and it predicted the energy levels of hydrogen

5. the Bohr Theory gave us a model of what the atom looks like and how it behaves.

• The analysis used in the Bohr Theory is also successful when applied to hydrogen-like atoms.

• An atom is said to be hydrogen-like when it contains only one electron. Examples are singly ionized helium, doubly ionized lithium, triply ionized beryllium, and so forth.

• The results of the Bohr theory for hydrogen can be extended to hydrogen- like atoms by substituting Ze2 for e2 in the hydrogen equations, where Z is the atomic number of the element.

28.18And 28.19Although many attempts were made to extend the Bohr Theory to

more complex, multi-electron atoms, the results were unsuccessful. Even today, only approximate methods are available for treating multi-electron atoms.

• EXAMPLE 28.2• Goal Apply the modified Bohr Theory to a hydrogen-like atom.• Problem Singly ionized helium, He+, a hydrogen-like system, has

one electron in the 1s orbit when the atom is in its ground state. Find (a) the energy of the system in the ground state in electron volts, and (b) the radius of the ground-state orbit.

• Strategy Part (a) requires substitution into the modified Bohr model, Equation 28.18. In part (b), modify Equation 28.9 for the radius of the Bohr orbits by replacing e 2 by Ze2, where Z is the number of protons in the nucleus.

• Solution• Find the energy of the system in the ground state. Write Equation

28.18 for the energies of a hydrogen-like system:

• Substitute the constants and convert to electron volts:

• Substitute Z =2 (the atomic number of helium) and n=1 to obtain the ground state energy:

Find the radius of the ground state. Generalize Equation 28.9 to a hydrogen-like atom by substituting Ze 2 for e 2:

For our case, n = 1 and Z = 2:Notice that for higher Z the energy of a hydrogen-

like atom is lower, which means that the electron is more tightly bound than in hydrogen. This results in a smaller atom, as seen in part (b).

• The number of protons in an atom’s nucleus is the atomic number for that particular element

• The same element may have different numbers of neutrons in its nucleus

• These slightly different kinds of the same elements are called isotopes