sapling question 8

[Karina Rodriguez 2H]

"A blue line is observed at 486.1 nm in the spectrum of atomic hydrogen. Determine the values of n for the beginning and ending energy levels of the electron during the emission of energy that leads to this spectral line." What steps do I need to take to find the n values? I'm unsure of what equation to refer to.

[Sahiti Annadata 3D]

Because it is an emission of energy we know that the electron transitioned from a higher to lower energy level. The lower energy level is n=2 because we know it visible light (486 nm). Then you can use rydberg’s formula to calculate the second energy level. The frequency is c/lambda (486 nm). Set this equal to R(1/n1^2 - 1/n2^2), and we know n2 is energy level 2. Hope this helps!

[Sebastian2I]

I followed these steps and ultimately arrived at a situation where I knew everything except the values of n1 and n2 inside the brackets. Is there no other way of solving other than guessing and checking? Am I missing some extremely obvious trick of algebra?

[Jason_Glass_2L]

I followed these steps and ultimately arrived at a situation where I knew everything except the values of n1 and n2 inside the brackets. Is there no other way of solving other than guessing and checking? Am I missing some extremely obvious trick of algebra?

Hello! Since this is a Balmer series (bc the wavelength is between 400 and 700), we know that the n1 value has to be 2. Since you already have the wavelength and have identified that n1=2 because of it being a Balmer series, you can solve for the n2. Hope this helps!

[Crystal Hsueh 2L]

You can tell what n-1 is when the problem states that it is a "blue light," meaning that it is visible light. This light is within the Balmer Series so you know right away that n-1 is 2. You can then find n-2 with the equation En= - (hR)/n^2

[Sebastian2I]

Thanks so much! This was definitely helpful!