NOTE: No Chem 1E03-01 students will have a formal tutorial during the week of October 11- 15. Solutions will be posted.
1. The McMaster student FM radio station CFMU broadcasts at a frequency of 93.3 MHz (MHz = 1 megaherz = 106 sec- 1).
a) What is the wavelength of this signal in meters?
nu = c / lambda
lambda = c / nu = (3.00 x 108 m s-1)/(96.3 x 106 s-1)
lambda = 3.22 m
b) What is the energy of one photon of this frequency?
E = h x nu = 6.62 x 10-34 J s x 9.33 x 107 s-1
E = 6.18 x 10-26 J
c) Compare the photon energy of part b with the energy of a photon of red light with wavelength 685 nm.
A red photon (its nu is about 10 15 s-1) has an energy of about E = h x nu = 6.62 x 10-34 J s x 1015 s-1 = 6.62 x 10-19 J. Its energy is about ten million times larger than the FM radio frequency of part a.
2. A cup containing 250 cm3 of water, initially at 25 oC, is warmed to 95 oC in a microwave oven that operates at a frequency of 2.45 GHz (GHz = 1 gigaherz = 109 sec-1). How many microwave photons of this frequency must be absorbed to warm the water? Assume that the heat capacity of the cup is negligible, compared to the heat capacity of the water.
Density of H2O = 1.00 g/cm3; thus, 250 cm3 H2O has a mass of 250 g
heat flow q = mass of H2O x Csp,H2O x delta T = 7.32 x 104 J
E of one microwave photon = h x nu = 6.62 x 10-34 J s x 2.45
x 109 s-1
E = 1.62 x 10-24 J/photon
# of photons required = 7.32 x 104 J/1.62 x 10-24
= 4.52 x 1028 photons
3. The longest-wavelength light that causes an electron to be emitted from 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?
KEelectron = E400 nm - E520 nm, where 520 nm corresponds to the minimum energy needed to remove the electron
KE = h x c x (1/lambda of 400 nm - 1/lambda of 520 nm)
KEelectron = [6.62 x 10-34 J s x 3.00 x 108 m/s]/[1.00 x 10-7 m x (1/4.00 - 1/5.20)]
KEelectron = 1.15 x 10-19 J/electron.
For one mole of electrons, KE = 1.15 x 10-19 J/electron x 6.022 x 1023 electrons/mole x 10-3 kJ/J
KE = 69.3 kJ/mol
4. Assume that you have graduated with a McMaster Engineering 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 x 10-19 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 associated with the highest energy possible photon on the planet is
E = h x c / lambda
= [6.62 x 10-34 J s x 3.00 x 108 m/s]/[5.40 x 10-7 m]
= 3.69 x 10-19 J
This photon is not energetic enough to eject an electron from the chosen metal. So the device will not work on the planet. You must choose another metal!
5. In the Bohr model for hydrogen atom,
En = -2.178 x 10-18 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
Ephoton required = difference in energy between E2 and E1
So we calculate what the photon energy must be to bring this about:
Ephoton = E2 - E1
= [(-2.178 x 10-18/4) - (-2.178 x 10-18/1)] J
Ephoton = 1.634 x 10-19 J
The photon of green light under consideration has an energy of
E = h x c / lambda
= [6.62 x 10-34 J s x 3.00 x 108 m/s]/[5.00 x 10-7 m]
= 3.97 x 10-19 J
So the green photon has too little energy to excite the electron from E1 to E2 (need a UV photon)
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.
a) 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 difference in energy levels. Since ninitial = 1, nfinal must be 2, so as to have the smallest delta E.
In problem 5 (above), we have calculated that this light must have an energy of 1.634 x 10-19 J
Since E = h x nu = h x c / lambda, lambda = hc/E
lambda = [6.62 x 10-34 J s x 3.00 x 108 m/s]/[1.634
x 10-18 J]
lambda = 1.21 x 10-7 m (or 121 nm)
b) What is the minimum amount of energy required to remove electrons from a mole of ground state hydrogen atoms?
To REMOVE an electron from H atom, we must consider n = 1 --> n = infinity. For one electron, this energy is h x nu. To remove an electron from N hydrogen atoms requires an energy of N x h x nu.
E = N x h x nu = N x (Einfinity - E1)
= 6.022 x 1023 x 2.178 x 10-18) J x [0 - (- 1)]
E = 1.31 x 106 J = 1.33 MJ (see p 333 in Kotz and Treichel)
7. Calculate the wavelength in nm for the longest wavelength emission transition seen in the Brackett series of hydrogen, namely where the final state corresponds to n = 4.
The Brackett series of lines are emission lines, all of which end in n = 4. The longest wavelength line in this series corresponds to the smallest energy gap, i.e., n = 5 --> n = 4.
E = h x nu = h x c / lambda; lambda = h x c / Ephoton
lambda = h x c /(En = 5 - En = 4)
6.62 x 10-34 J s x 3.00 x 108 m/s
[(-2.178 x 10-18)/25 - (-2.178 x 10-18)/16)]J
lambda = 4.05 x 10-6 m (4050 nm)
8. 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. Show your calculations.
Ephoton = h x c / lambda
6.62 x 10-34 J s x 3.00 x 108 m/s
2.625 x 10-6m
Ephoton = 7.568 x 10-20J
This is a Brackett series line, nlower = 4.
Therefore Delta Ehydrogen = 7.568 x 10-20J = -2.178 x 10-18 (1/n2 - 1/16)J
7.568 x 10-20 J - 13.61 x 10-20 J = -2.178 x 10-18 J/(n2)
n2 = 2.178 x 10-18/6.04 x 10-20 = 36.06
Therefore n = 6 (6 --> 4 transition)
9. 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.
We need to determine those transitons that involve LESS energy than the 3 --> 1 transition. Therefore the 4 --> 1 and 5--> 1 transitions are eliminated, since they require more energy. 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.