Spring Chemistry Homework #3 Answer Key

Introduction to Quantum Chemistry

Oxtoby, Nachtrieb and Gillis: Chapter 15:

5.

11. The work function is the amount of energy that a single photon must carry to eject a photon from the surface of a metal. If the work function for cesium is 3.43 × 10^-19 J, then,

 

This wavelength corresponds to yellow light. So wavelengths of light that have a wavelength of 580 nm or less will result in a photocurrent from cesium. This includes green, blue and violet light, but not orange or red.

For selenium, the work function is 9.5 × 10^-19 J. This requires light in the ultra-violet portion of the spectrum to produce a photocurrent.

 

25. The radius of "hydrogen-like" atoms is given by:

where:
n: Bohr quantum level
e_o: permitivity of free space
Z: atomic number (magnitude of nuclear charge)
e: electron charge
m: mass of the electron

For an electron in the third quantum level of a B⁴⁺ ion, we have:

 

The energy of a hydrogen-like atom is given by:

where R_H is the Rydberg constant (2.18 × 10^-18 J). So, for the B⁴⁺ ion with an electron in the n = 3 quantum level:

For an n = 3 to n = 2, we have DE = E₂ - E₃. Or,

This is the energy change for the atom itself - it is negative because the atom went from an excited, or "high energy", state to a lower energy state. The energy "lost" was emitted as a photon with the following frequency:

To remove an electron a B⁴⁺ ion in the n = 3 state would require:

So, for 1 mol of such ions,

 

a) For an electron moving at 1.00 × 10³ m s^-1:

 

b) For a proton moving at the same speed:

 

c) For a baseball thrown by Sidd Finch,

 

 

To find a possible transition of the C⁵⁺ ion that would a photon of this energy, we can do the following:

The value 0.005 is close to 0.00478, which corresponds to the term on the left, when n_f = 7 and n_i = 8.

The actual wavelength for an n = 8 to n = 7 transition in the C⁵⁺ ion has a wavelength of 530 nm, which is still in the green portion of the spectrum.

90. The energy required to raise the water temperature is:

The energy of one photon of the given frequency is:

E = hf = (6.636 × 10^-34 J s) (2.45 × 10⁹ s^-1) = 1.63 × 10^-24 J

So, the total number of photons required is:

 

2.    a) The n = 1 has the ground state configuration.
       b) The electron is moving faster in the n = 1 state.
       c) The n = 4 state has the larger radius.
       d) The n = 1 state has the lower potential energy
       e) The n = 1 state has the larger ionization energy.

5.

E = hf = (6.636 × 10^-34 J s) (0.66 × 10¹⁵ s^-1) = 4.38 × 10^-19 J
 

Wavenumbers are a means of expressing frequency. It corresponds to the number of cyclces per centimeter, or, cm^-1.

 

So the energy released is + 3.03 × 10^-19 J.

18. Three lines are observed because there are three transitions possible. These are:

n = 3 to n = 2
n = 3 to n = 1

n = 2 to n = 1

The shortest wavelength of light corresponds to the n = 3 to n = 1. This has a frequency of :

The energy difference between the first and third levels is:
E = hf = (3.00 × 10¹⁵ s^-1) (6.626 × 10^-34 J s) = 1.99 × 10^-18 J
 

For the 125 nm light:

The energy difference between these levels is:
E = hf = (2.40 × 10¹⁵ s^-1) (6.626 × 10^-34 J s) = 1.59 × 10^-18 J
 

For the 500 nm wavelength light:

The energy difference between these levels is:
E = hf = (6.00 × 10¹⁴ s^-1) (6.626 × 10^-34 J s) = 3.98 × 10^-19 J