Excitation of hydrogen atom by visible light Question statement: 

A beam of light having wavelengths distributed uniformly between 450nm and 550nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?

When they say least intensity, I assume that a particular wavelength of light will be absorbed to excite the electron, and hence there will be less intensity of that wavelength.
I calculated the energy associated with the beam to range from approx. 2.45eV to 2.75eV.
Considering that the minimum energy required to make a transition from ground state to a higher energy state is 10.2eV, my conclusion was that each wavelength of light would have the same intensity. 
However, the answer given is wavelength = 487nm. On referring an online solution, I found that they took initial principal quantum number (n) of the electron as 2 instead of 1. Is this correct? They justified this by saying that the wavelengths of the incoming beam lie in the visible region.
Don't the electrons of a H2 molecule have a principal quantum number of 1 regardless of the region of incident light?
 A: I have doubts about the relevance of your statement "the minimum energy required to make a transition from ground state to a higher energy state is 10.2eV" for molecular hydrogen, as there are vibrational and rotational energy levels of the hydrogen molecule and at least quadrupole transitions among them (http://www.eso.org/~tstanke/thesis/chap2_10.html).
A: The answer is correct if one presupposes that the H atom is already excited to  higher levels including the level of n=2, and gets a further excitation. I suppose because of  collisions of the atoms statistically, some electrons may be at the n=2 level.
I think the question gives a narrow band choice so only one level can be active, though I do not see the educational use of the question, unless it is for a specific space region or something like that.
A: 
Don't the electrons of a $\mathrm{H}_2$ molecule have a principal quantum number of 1 regardless of the region of incident light?

Hydrogen atoms in the ground state have principle quantum number 1, but there is no requirement that hydrogen atoms in interstellar space remain in the ground state (a statement abundantly justified by the fact that we observe Balmer series absorption lines in starlight that has passed through nebulae). There is a lot of energetic stuff going on out there (cosmic rays, in particular, are ubiquitous), that can excite or ionize atoms.
Moreover, you ask about hydrogen molecules which have their own spectrum different from that of atomic hydrogen which is not well described by the Rydberg formula.
A: Atomic hydrogen has this emission line  in the Balmer series at 486 nm. But you are right, this is not an absorption line, as there are no atoms with an electron in the $2s$ state in a normal sample of hydrogen. That question was mistaken. 
But in space, in the outer atmospheres of stars, when the temperature is high enough (but not too high to ionize all atoms), this gives rise to the H$\beta$ absorption line. Stars where this absorption line is strong have spectral classification A. They have a temperature of about 10 000 K.
I can reasonably assume that the question was about atomic hydrogen. But also H$_2$ is transparent in the visible.
A: The question is kind of gloriously ambiguous, so it's hard to draw a hard line between right answers and wrong answers here. However, the short of it is this:


*

*as far as the absorption of radiation, the phrase "a sample of hydrogen gas" is pretty meaningless if it doesn't mention its temperature.


If the gas is cold, then it will be in the form of molecular hydrogen, for which none of the spectral series of atomic hydrogen apply $-$ and indeed, molecular hydrogen has rather weak radiative interactions in the visible because it is symmetric.
Conversely, if the gas is extremely hot, then it will be completely ionized; in those conditions it is opaque to radiation of all frequencies (with a preference for the proton and electron plasma frequencies), and it does not give any special treatment to the spectral series of atomic hydrogen.
Somewhere in the middle, there is a temperature range where some or all of the hydrogen dissociates into atomic hydrogen. At the lower end of this range, this will overwhelmingly just populate the electronic ground states of those hydrogen atoms, but as the temperature increases, more and more of them will be found in excited electronic states. In that situation, if you shine broadband light in the visible range onto your sample, some of those excited atoms will be in the $n=2$ states and they will be happy to absorb that visible light.
So, with that in mind: take that question as one that wasn't very well written, and just move on.
