What Temperature is Needed for Hydrogen Gas to Produce a 656.2nm Emission Line?

[indie452]

the question states that the emission spectrum of hydrogen contains a line with wavelength 656.2nm. and we need to approx find out what temp the H gas needs to have to be heated before this line appers in the spectrum

i have worked out that for this wavelength the electron mov from level 3 to 2, and that the energy between these two levels is ~1.89ev (3.03E-19 J)

im not sure whether i can use the internal energy formula = 1/2(RfnT) = 1/2(3RT)
or the weins law for blackbodies => lamda = 2.898E-3/T

 

[nasu]

If you are looking for an aproximation (oder of magnitude) you can just use kBT for average thermal energy per molecule.
kB is Boltzmann constant.

 

[ideasrule]

im not sure whether i can use the internal energy formula = 1/2(RfnT) = 1/2(3RT)
or the weins law for blackbodies => lamda = 2.898E-3/T

I don't know how you plan to use lamda = 2.898E-3/T, but you can certainly use 1/2(3RT). 1/2(3RT) is the internal energy that each mole of gas has; 1/2(3kT) is the internal energy of each molecule.

 

[indie452]

but if i use 1/2(3kT) i get for the temp = 14637K this seems too big as H-alpha lines can be seen in the sun and that has a temp of ~5000K which is about 3 times smaller, but if i use 2.898E-3/T i get 4416K which seems more reasonable, but i would have thought the first way would give a reasonable answer. if it i correct then what does the 14637K represent?

 

[nasu]

Wien's law is for a black body radiation. It does not really apply to the question you have here.

The question is about just an estimate of the order of magnitude. It does not really matter if you take the factor 3/2 in front of the energy. That energy is the AVERAGE thermal energy.
Even at lower temperatures when the average is lower, there are molecules with higher energies that can excite the line considered.
So you get something like 10,000 K. The Sun has 6000 K. Is the same order of magnitude.
That's all you can expect from this estimate. For more accurate calculation you need to look at the actual distribution of the energies and define what exactly is required to observe the line in the spectrum.