Balmer Series Quiz Solution and comments

Let's start by talking about energy units and then doing parts 1) and 4).
We use f=c/wavelength for 1), and then E=hf for 4). Almost everyone got 1), but for many people there was some confusion regarding units of energy for 4). Electron Volts (eV) is a unit of energy that is not the same as Joules. 1 eV = 1.6 x 10^-19 Joules, so an eV is a much smaller unit of energy and one that works well for atoms in the sense that it gets rid of the large negative exponents that occur if one uses Joules.

eV has its origin in the potential energy of an electron, so it makes some sense that it is close to the right size for electrons in atoms; Joules, on the other hand, has its origin in the kinetic energy of a 2 kg mass moving at 1 m/s, and is way off (too big) for atom energies.

Now to the problem:

f=c/lamda = (3 x 10^8 m/s) /656 x 10^-9 m = (3/656) x 10^17 sec-1 = 4.57 x 10^14 Hz .

Getting the exponent right in the 3rd step is critical. Exponents are very important!
For the others you should get: 6.17, 6.91 and 7.31 x 10^14 Hz, respectively. (All are multiplied by 10^14. All are in cycles per second, which is the same as Hz and sec-1.)

4) For 4, you multiply each one by h=4.14 x 10^-15 eV-sec to get the energy in eV.

E= h f = 4.14 x 10^-15 eV-sec 4.57 x 10^14 Hz = 18.9 x 10^-1 eV = 1.89 eV.

[Note that this is a nice, friendly number. No exponents.]

The other ones are: 2.56 eV, 2.86 eV and 3.02 eV.
These photon energies correspond to something like: 1.89 eV (red or maybe orange), 2.56 eV (greenish blue?), 2.86 eV (blue) and 3.02 (violet).

The answer is very different in Joules. For example, for the first frequency you would get about 3 x 10^-19 Joules.

We could just forget about eV if people would rather work with Joules. There is sort of a tradeoff between the extra work and confusion of learning a new unit for energy, and the benefit of using a unit (eV) that is more atom friendly. Please let me know your preference via comments here.