Balmer Series and Hydrogen?

Question

Hi I just did a lab for Balmer series. For v bar one of my values was 2.67 x 10^7. I am using the equation: v bar = R (0.25 - (1/n^2)). I am supposed to solve for n but I don't think I am getting the right answer. I don't think n should be negative. Can someone explain this to me?

Answer 1

The energy per photon released in the transition of an electron from a higher energy state to a lower energy state is given as,
E = hv,
Where E is the energy, h is Planck’s constant, and v is the frequency.
Note, we are using h, not h-bar.

The inverse wavelength, k, corresponding to the light emitted during the can be found as,
k = R(1/(n2)^2 – 1/(n1)^2)
Where n2, in the Balmer series, is always 2, n1 is the initial energy level of the electron (n1 = 3, 4, 5, 6, …..), and R is the Rydberg constant (=10,973,735.3 m−1, for an infinitely massive nucleus).
λ = c / v
k = 1 / λ
k = v / c

Your goal is to find n1. You know that n2 will always be 2 for the Balmer series, you know the Rydberg constant (or you are assumed to know it), and you know the frequency (and thus also the wavelength) of the light emitted. You know all you need to in order to solve for n1 using the second equation.

All of the above is great, in theory, but when you put it into practice with the information you stated…it is not going to work.
You must have read / typed something incorrectly because light with the frequency you stated is not within the range of the Balmer series, it is too low of an energy. It would make a little more sense if the number you have for the frequency was really the wavelength of the light, but even this is outside of the range of the Balmer series (about 350 to 650 nanometers).
In order to solve this problem you would need to use another series (other than the Balmer). In this case, the final quantum number of the electron will not be 2, it will likely be much larger since at larger values of n, the difference between the energy levels shrinks and it allows for smaller transitions of energy.

Go back and check you numbers for the frequency and/or wavelength.

Answer 2

the hydrogen atom needs a lot of energy to excite its electrons enough to emit photons and produce light, but not all of the energy put into the atom goes into the electrons, most of it becomes heat. since the majority of the energy you are putting into the gas is used to heat it up, obviously it becomes very hot, and since it is a gas it expands. And because it is in a glass container of unchanging volume, it is unable to expand and the pressure increases more and more as it is heated up. The gas needs to be at a lower pressure to prevent it from breaking the glass as it is heated and the pressure increases. If the gas started at a high pressure, it would be easier to have enough pressure to break the glass, which would be extremely dangerous because of both the resulting shards, and the highly combustible hydrogen gas. also at a lower pressure there would be less hydrogen atoms in the glass to excite, whereas at high pressure more atoms are present which require more energy to excite, so a current from a conventional wall outlet might not have enough energy to light up the atoms, e.g. lighting a cubic centimeter of hydrogen versus a cubic liter at the same temperature and pressure.