1E03
TUTORIAL CHEM #4
Solutions
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1. The McMaster student FM radio station CFMU broadcasts at a frequency of 93.3
MHz (MHz = 1 megahertz = 106 s1).
a) What is the wavelength of this signal in meters?
= c / = (2.9979 108 m s-1) / (93.3 106 s1) = 3.213 m = 3.21 m
b) What is the energy of one photon of this frequency?
E = h = 6.626 1034 Js 9.33 107 s1 = 6.182 1026 J = 6.18 1026 J
c) Compare the photon energy of part b with the energy of a photon of red light
with wavelength 685 nm.
E = h, so determine and use to solve for E (or use E = hc/).
= c / = (2.9979 108 m s-1) / (685 109 m) = 4.376 1014 s1
A red photon (its frequency is about 1014 s1) has an energy of
E = h = 6.626 1034 Js 4.376 1014 s1 = 2.899 1019 J = 2.90 1019 J.
Its energy is about five million times larger than the FM radio frequency of part a.
2. Sodium vapour lamps used for street lighting emit two yellow lines that are part
of the atomic spectrum of sodium. One of the lines has a wavelength of 589.0 nm,
and the other has a wavelength of 589.6 nm.
a) Express each of these wavelengths in meters.
589.0 nm 1 m/109 nm = 5.890 107 m
589.6 nm 1 m/109 nm = 5.896 107 m
b) What is the frequency of each of these lines?
= c / = [2.998 108 m/s]/5.890 107 m = 5.0898 1014 s1 = 5.090 1014 s1 (or Hz)
= c / = [2.998 108 m/s]/5.896 107 m = 5.0846 1014 s1 = 5.085 1014 s1 (or Hz)
c) How much more energy, in joules, does a photon of 589.0 nm possess, as
compared to a photon of 589.6 nm?
E = h1 - h2 = h(1 - 2) = 6.626 1034 Js (5.0898 5.0846) 1014 s1
E = 6.626 1034 Js (0.0052) 1014 s1 = 3.4 1022 J = 3 1022 J
(Only 1 sig. fig. in frequency difference, so 1 sig. fig. in energy difference)
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CHEM 1E03
TUTORIAL #4
Solutions
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3. The longest-wavelength (i.e. lowest energy) light that causes an electron to be emitted
from a gaseous lithium atom is 520. nm. Gaseous lithium atoms are irradiated with light
of wavelength 400. nm. What is the kinetic energy of the emitted electrons, in kJ/mol?
h = hc/ = KE + BE, where KE = kinetic energy and BE = binding energy of the electron to
the atom (or threshold energy). From the threshold wavelength, BE = hc/520nm.
KE(electron) = E[400 nm] E[520 nm] = hc (1/400. nm - 1/520. nm)
= [6.626 1034 Js 2.9979 108 m/s] [(1/4.00 107m) (1/5.20 107m)]
= 1.146 1019 J/electron.
For one mole of electrons,
KE = (1.146 1019 J/electron) (6.022 1023 electrons/mole) (103 kJ / J) = 69.0 kJ/mol
4. Assume that you have graduated with a McMaster degree, and that you are designing
a space probe to land on a distant planet. You wish to use a switch that works by the
photoelectric effect. The metal that you wish to use in your device
requires 6.7 1019 J/atom to eject an electron. You know that the atmosphere of the
planet on which your device must work filters out all light of wavelengths shorter than
540 nm. Will your device work on the planet? Why or why not?
The energy of the shortest wavelength (highest energy) photon possible on the planet is:
E = hc / = [6.62 1034 Js 2.9979 108 m/s] / [5.40 107 m] = 3.69 1019 J
This photon is not energetic enough to eject an electron from the chosen metal. Any photon
with longer wavelength will have less energy. So the device will not work on the planet, as
the required range of wavelengths will be filtered out. You must choose another metal!
5. In the Bohr model for hydrogen atom, En = 2.178 1018 J/n2 .
Calculate whether a photon of green light, of wavelength 500. nm, has enough
energy to excite the electron in the hydrogen atom from n = 1 to n = 2.
Here we are concerned with the transition E1 E2
E[photon] required = difference in energy between E2 and E1
So we calculate what the photon energy must be to bring this about:
E[photon] E2 = E1 = [(2.178 1018 J/ 4) (2.178 1018 J/1)] = 1.634 1018 J
The photon of green light under consideration has an energy of E = hc /
E = [6.626 1034 J s 2.9979 108 m/s]/[5.00 107 m] = 3.97 1019 J
So the green photon has too little energy to excite the electron from E1 to E2.
(We need a UV photon for any transition from the ground state.)
Note that photons of higher energy ( > 1.6 1018J) cannot cause this transition either.
Only photons of exactly the right energy can be absorbed when moving from one bound
state (n < infinity) to another.
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CHEM 1E03
TUTORIAL #4
Solutions
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6. When the hydrogen atom is at relatively low temperatures, its electron is in the
lowest energy level, n = 1. This is called the ground state.
What is the longest wavelength of radiation that can be absorbed by such an atom?
The longest wavelength of light that can be absorbed corresponds to the smallest
frequency and therefore the smallest difference in energy levels.
Since n[initial] = 1, n[final] must be 2, so as to have the smallest energy change.
In problem 5 (above) we calculated that each photon of this light must have
1.6335 1018 J of energy.
Since for a photon E = h = hc / , then = hc/E
= [6.626 1034 Js 2.9979 108 m/s] / [1.6335 1018 J] = 1.216 107 m = 122 nm
7. The Brackett series of emission lines from atomic hydrogen occurs in the far
infrared region. One of the lines has a wavelength of 2625 nm. Determine the values for
the quantum number n for the two energy levels involved
in the transition.
For an emission process, Ei must be larger than Ef and ni is larger than nf.
hc/ = Ei - Ef = -K/ni2 - (-K/nf2) = K (1/nf2 - 1/ni2) = (2.178 1018 J) (1/nf2 - 1/ni2)
Hence, for the transition with = 2625 109 m, (1/nf2 - 1/ni2) = 0.03477 - equation (1).
A transition in the far infrared region must correspond to nf = 3,4,5. Larger values of nf
would yield negative values of ni in equation (1), which is impossible.
nf = 3 ni = 3.6 not possible
nf = 4 ni = 6.0
nf = 5 ni = 13.8 not possible
The emission corresponds to the n = 6 n = 4 transition.
8. The n = 3 to n = 1 emission line for atomic hydrogen occurs in the UV region (it is
a member of the Lyman series).
Without doing any calculations, decide which of the following emission lines for atomic
hydrogen occur at longer wavelengths than this line.
a) n = 4 n = 2;
b) n = 4 n = 1;
c) n = 5 n = 1;
d) n = 5 n = 3.
A preliminary ordering can be determined by looking at the lowest state in each
transition. The lower this state the higher the energy of the transition (see fig 8.14 of
Petrucci).
The 4 2 transition is a VISIBLE transition (Balmer series) and requires less
energy than the UV transition. Likewise the 5 3 transition, in the infrared region,
requires less energy than the UV transition.
Within those transitions ending at n =1, the higher the initial state the higher energy is
released in the transition. Therefore the 4 1 and 5 1 transitions are eliminated,
since they require more energy.
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CHEM 1E03
TUTORIAL #4
Solutions
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9. The energy needed to dissociate a chlorine molecule into chlorine atoms (the
bond energy) is 243 kJ/mol. What is the longest wavelength of light that can break
the Cl- Cl bond? What is the colour of light of this wavelength?
E = 243 kJ/mol (1 mol/ 6.022 1023 molecules) (1000 J/1 kJ) = 4.035 1019 J/molecule.
One photon is needed to dissociate one Cl2 molecule. Its frequency, , is
= E/h = 4.035 1019 J / 6.626 1034 Js = 6.09 1014 s1 (or Hz)
= c/ = [3.00 108 m/s] / [6.09 1014 s1] = 4.93 107 m = 493 nm
This corresponds to green light. Any light of higher energy can also be used, since the
extra energy becomes kinetic energy of the separated Cl atoms.
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