Quick help with a chemistry conundrum...?

Question

These are for my chem. homework and I was wondering if I could get a few hints/formulas as how to solve these. I think I have the first one, but the second one is leaving me clueless...
help would be GREATLY appreciated!

1. What is the wavelength of light (in meters) whose frequency in the FM radiowave region is 103.9 MHz?

...and the really tough one...

2. The absorption of a photon of wavelength 6.71x10^2 nm causes an electron in a certain molecule to go from one energy level to a state of higher energy. Absorption of energy in this range causes an electron in the valence shell of a molecule to move to a higher level. If the higher electronic level has an energy of 4.840x10^-19 J, what is the energy of the lower electronic state?

Answer 1

For the first, the formula relating wavelength (L) and frequency (f) of a light wave is

L = c/f

where c is the velocity of light (3*10^8 m/sec).

The energy equivalent of a light photon is given by

e = h*f

where h = Planck's constant, 6.626*10^-34 joule-sec.

Use the first formula to find f from the wavelength, then the above formula to get the energy.

The energy of the lower state is then the energy of the higher state minus the photon energy from the above calculation.

Answer 2

For part 1 (the easy part): the speed of light and radio
waves and all electromagnetic radiation is denoted as c
by physicists, and is about 300 million meters per second.
The wavelength is found by dividing the speed (c) by the
frequency; in this case, 103,900,000 cycles per second.
300 million divided by 103.9 million = 2.887 meters
The wavelength of the FM radiation is 2.887 meters.
Now comes the hard part (2) but it isn't that bad.
The energy contained in a photon is directly proportional to
its frequency f, and the constant of proportionality is called
Planck's constant, abbreviated h by physicists. Its value
is 6.625* 10 to the -34 joule-seconds. Multiply the frequency
by h and get the energy of the photon in joules.
Here you are given the wavelength of the photon and not
the frequency; since E = hf, E also equals hc/w, the wave
length. h and c are constants; their product is 3 * 10 **8
times 6.625 * 10**-34 = 19.875 * 10**-26
Divide that by the wavelength in meters, or 6.71 * 10**-7
meters, and you get the energy of the photon as
2.962 * 10**-19 joule. This is the amount of energy added
by the photon when it hit the electron and raised its energy
to the higher level, given as 4.84 * 10**-19 joule.
The lower energy level is found by subtracting the energy
added by the photon from the final energy level. We get
(4.84 - 2.962) * 10**-19 joule, or 1.878 * 10**-19 joule
as the energy in the lower level.
Simple, huh?
If this is freshman chemistry, you will have fun!

Answer 3

c = wavelength*frequency delE = hf = hc/wavelength
delE = E2-E1
1. What is the wavelength of light (in meters) whose frequency in the FM radiowave region is 103.9 MHz?
wavelength = c/frequency = 3.000x10^8/(103.9 x 10^6) m
= 2.887m

2. The absorption of a photon of wavelength 6.71x10^2 nm causes an electron in a certain molecule to go from one energy level to a state of higher energy. Absorption of energy in this range causes an electron in the valence shell of a molecule to move to a higher level. If the higher electronic level has an energy of 4.840x10^-19 J, what is the energy of the lower electronic state?
delE = h*3.00x10^8/(6.71x10^2x10^-9) J
= 6.626x10^-34* 0.44709x10^15 J
=2.962 x10^-19J

Elow = Ehi - delE
= (4.840 - 2.962)x10^-19J
= 1.878 x10^-19J

Answer 4

The answer to that can be found in the set up to the 17th problem on the May, 2003 past paper.

Answer 5

103.9 Mhz