Plz Help! (Atomic Spectrum/wavelengths)

[CDerhammer]

Calculate the wavenumber (v=1/lambda) and the wavelength of the first transition in the visible region of the atomic spectrum of hydrogen.

I KNOW THE ANSWER..... 656.3 nm. I just don't have a clue how they got that answer, I do know that I had to use Rydberg's Constant (1.097X10^7 m ^-1) but I have no idea what to do or how to do it. Thank you anyone who wants to take time and help me. And I don't know if this will mean anything but it is Inorganic Chem 430.

[phision]

Calculate the wavenumber (v=1/lambda) and the wavelength of the first transition in the visible region of the atomic spectrum of hydrogen.

I KNOW THE ANSWER..... 656.3 nm. I just don't have a clue how they got that answer, I do know that I had to use Rydberg's Constant (1.097X10^7 m ^-1) but I have no idea what to do or how to do it. Thank you anyone who wants to take time and help me. And I don't know if this will mean anything but it is Inorganic Chem 430.

 

The terminology in your question may be lecturer/ tutor specific but as I understand it the 656nm emition is the visible emition with the lowest energy (H-alpha) and the Rydberg constant is assoiciatated with the ionisation energy. Would it be possible to clarify the question specifics and then re-post the whole question?

[phision]

Some time on Google has let me understand that: the inverse of the wavelength(λ),of the emition, is equal to the Rydberg constant multiplied by the difference of the inverse square of of the wave numbers, of the energy levels, that the transition has occured between(i.e. 2 and 3).

 

So: 1/λ=R_∞((1/2²)-(1/3²))

 

∴ 1/λ=10 973 731.568 539 m^-1x(5/36)=1524129.384519305...m^-1 ( I think you called this the "wavenumber (v=1/lambda)" in the original post)

∴ λ=6.5611227639665875623523370390403x10⁷m OR 656.1nm