CH 115 Fall 2015
Worksheet 2
1. The Aurora Borealis is a night display in the Northern latitudes caused by ionizing
radiation interacting with the Earth's magnetic field and the upper atmosphere. The
distinctive green color is caused by the interaction of the radiation with oxygen and has a
wavelength of 5577 angstroms. What is the frequency of this light? How much energy is
released?
Given wavelength, we are asked to find frequency. The equation that relates the two
of these is c (speed of light) = wavelength (lambda) * frequency (nu). The speed of
light is a constant that we approximate as 3.0 x 108 m/s. Given the wavelength, we
can now solve for the frequency.
1 m = 1010 angstroms  5577 A = 5577 x 10-10 m  5.577 x 10-7 m
frequency = 3.0 x 108 / 5.577 x 10-7 = 5.379236148 x 1014 1/s (Hertz, Hz)  remember
NOT to round until the end of a problem as it will mess up your significant figures
(we’ll talk more about this next week)
Now, we are asked to find the amount of energy released. The equation we want to
use here is E (energy) = h (Planck’s constant) * frequency (nu). Planck’s constant is
always 6.626 * 10-34 Js. We have the frequency we figured out above, so we can
multiply Planck’s constant times that to get the amount of energy in Joules that was
released.
E = (6.626 x 10-34) x (5.379236147 x 1014) = 3.564 x 1019 J (now you can round)
NOTE: Make sure ALL of your units match up and cancel out before you do ANY
math. The most common mistake in doing these types of problems is just using
the numbers as given, rather than converting them to the units you need!
2. If it takes 3.36 x 10-19 J of energy to eject an electron from the surface of a certain metal,
calculate the longest possible wavelength, in nanometers, of light that can ionize the
metal.
Working backwards from energy, we should start with the equation E = hv, and
solve for the frequency.
E = hv  v = E / h (plug and chug) = 3.36 x 10-19 J / 6.626 x 10-34 Js = 5.070932689 x
1014 1/s (DO NOT ROUND YET, ONLY AT THE END OF A PROBLEM)
Now that we have frequency, we can use the equation c = wavelength x frequency to
get the wavelength.
Wavelength = c / frequency = 2.99 x 108 m/s / 5.070932689 x 1014 1/s = 5.89635119 x
10-7 m
The question asks for the wavelength in nanometers, so we have to convert our
answer from meters to nanometers.
CH 115 Fall 2015
Worksheet 2
5.89635119 x 10-7 m x 109 nm = 589.64 nm (feel free to round now!)
3. Describe the observed line spectra graph. Include observations about emitting and
absorbing energy and relative energy level transitions.
This is the simplest version I could find – there’s probably a better one in your book so
check it out!
A line spectra graph or a Bohr diagram describes energy levels of the electrons present
in the atom. Energy level is represented by the n on the right side of the diagram and as
we increase n, we increase in energy. If an electron moves up the diagram (say from 1 to
2) it is ABSORBING energy. If it moves down the diagram (from 2 to 1) it is
EMITTING or RELEASING energy. Notice that the gaps between energy levels are
larger at the bottom and smaller at the top. It becomes easier (takes less energy) to shift
energy levels at the top than at the bottom. In other words, it takes a whole lot of energy
to move from n=1 to 2, while it takes only a little energy to move from n=3 to 4.
4. What are the four quantum numbers used to describe an electron and what about the
electron do they describe?
Quantum numbers are used to describe aspects of an electron present in a
particular atom. The term “quantum” comes from the word quantized, which
means a discrete unit or packet. So basically electrons are present in discrete units
and are only allowed to be present in these specific locations/energy levels. There are
four quantum numbers used in general chemistry to describe characteristics of an
electron.
Principal quantum number (abbreviated n) – determines size and RELATIVE
ENERGY (what energy level the electron is in
- Can be any positive integer (from 1 to infinity)
- Also corresponds to the row of the element in the periodic table, ex. An
electron in the lithium atom has a principal quantum number of n = 3
- As you increase the number, the energy increases
CH 115 Fall 2015
Worksheet 2
Subsidiary quantum number (abbreviated l) – determines shape or the subshell the
electron is present in (each subshell has a characteristic shape)
- Can be any number from 0 to (n-1)  for an electron with n = 3, l can be
0, 1, or 2
- Each subshell can also be represented by a letter (0 = s, 1 = p, 2 = d, 3 = f)
 corresponds to blocks on the periodic table
- Shapes: s = sphere, p = dumbbell, d = clover leaf or dumbbell inside a
doughnut hole, etc. (look in your book if you don’t understand this)
Magnetic quantum number (abbreviated ml) – determines the orientation of the
subshell that the electron is in or the orbital (each subshell will have a particular
number of orbitals)
- Can be any number from –l to +l  for an electron with l = 1, ml can be 1, 0, or 1 (this subshell has THREE possible orientations)
- Generally, s has 1 orientation, p has 3, d has 5, and f has 7
Spin quantum number (abbreviated ms) – determines which direction the electron is
spinning
- Only two possible spins (-1/2 or +1/2)
- There are two electrons in every orientation or orbital  each of these
electrons has a different spin (one is + and one is – ALWAYS)
5. List all the possible combinations of the four quantum numbers (n, l, ml, ms) for n=3.
(3, 0, 0, + or – ½)
(3, 1, -1, + or – ½)
(3, 1, 0, + or – ½)
(3, 1, 1, + or – ½)
(3, 2, -2, + or – ½)
(3, 2, -1, + or – ½)
(3, 2, 0, + or – ½)
(3, 2, 1, + or – ½)
(3, 2, 2, + or – ½)
You should write out the + ½ and the – ½ as separate sets of quantum numbers, so
there really should be 18 total!
6. What orbitals are described by the following sets of quantum numbers?
a). (4, 3, -2, ½) 4f
b). (3, 2, 0, -1/2) 3d
c). (1, 0, 0, -1/2) 1s
d). (2, 1, 1, ½) 2p
e). (5, 4, 0, ½) OMIT THIS QUESTION
f). (2, 0, 0, ½) 2s
7. Which set of four quantum numbers not possible? C (l always has to be less than n)
a. n = 4 l = 2 m = -1 s = +1/2
CH 115 Fall 2015
Worksheet 2
b. n = 3 l = 0 m = 0 s = -1/2
c. n = 2 l = 2 m = 1 s = ½
d. n = 5 l = 3 m = -2 s = -1/2
8. Which type of orbital is not possible? A
a. 3f
b. 6s
c. 2p
d. 4p