CH 115 Fall 2015Worksheet 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 * 10-34Js. 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!

  1. 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 10-34Js = 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.

5.89635119 x 10-7 m x 109 nm = 589.64 nm (feel free to round now!)

  1. 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.

  1. 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

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)

  1. 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!

  1. 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

  1. 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

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

  1. Which type of orbital is not possible? A

a. 3fb. 6sc. 2pd. 4p