1.15....Where did I go wrong? [ENDORSED]

[Chin_Alyssa_3I]

1.15 In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line.

[sol] Since it is a Lyman series, I know that nf (n final) is 1. I also know that {ΔE= Ef - Ei}.

Since {ΔE = hc/λ}...

then {hc/λ = Ef - Ei}

1.99e-25/1.026e-7 = (-hR/1^2) - (-hR/ni^2)

1.94e-18 = -2.18e-18 + (-2.18e-18/ni^2)

4.12e-18 ni^2 = -2.18e-18

ni^2 = .529

ni = .727 (???)

I know that since energy is quantized, my answer needs to be a whole number greater than 1 (the final energy level) so where did I go wrong?

[Amy Ko 3C]

ΔE= Ef - Ei is the correct approach to this problem. However, ΔE is negative when electron is moving down in energy level, meaning n final < n initial.

[Mizuno_Mikaela_1D]

I know Amy answered this but I figured I would show out all the math for anyone else that needs it.
Note: I changed the ΔE so that it was negative since the wavelength is being emitted. You would get the same energy levels if you had a positive ΔE, but ni would have to be 1 instead of nf since when ΔE is positive, the electron is going from a lower energy level to a higher energy level but that would be when a wavelength of light is absorbed I believe.

1.15 In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line.

Step 1: Find ΔE
c=λv E=hv
E=hc/λ
E= (6.62608*10^-34Js)(2.99792*10^8m/s)/(102.6*10^-9m)
E=1.93610699*10^-18J
Since the light is being emitted, the electron is moving down from a higher energy level to a lower energy level which means the ΔE must be negative. Therefore..
ΔE= - 1.93610699*10^-18J

Step 2: Find Ef
Since the wavelength of light emitted is in the ultraviolet spectrum of atomic hydrogen, we know it is in the Lyman series which means nf(n final)=1.
Ef= - hR/nf^2
Ef= - (6.62608*10^-34Js)(3.28984*10^15Hz)/(1^2)
Ef= - 2.1798743*10^-18J

Step 3: Find Ei
ΔE=Ef-Ei
Ei=Ef-ΔE
Ei= - 2.1798743*10^-18J - (- 1.93610699*10^-18J)
Ei= - 2.4376731*10^-19J

Step 4: Solve for ni
Ei= - hR/ni^2
(ni^2)(Ei)= - hR
ni^2= - hR/Ei
ni^2= - (6.62608*10^-34Js)(3.28984*10^15Hz)/(- 2.4376731*10^-19J)
Take the square root of both sides:
ni = 3

During the emission of energy of the 102.6nm wavelength, the electron moves from an energy level of n=3 down to an energy level of n=1.

[Nadine_El Fawal_3M]

I got the same answer and agree that nf=1 and ni=3. However; the solution manual says that the electron goes from n=1 to n=3 and not vice versa. even though the question specifically says that light is being emitted (hence the e- is going from a high E-level to a lower one) and that it's an ultraviolet light (i.e: Laymen's series, meaning that the final E-level has to be 1) Is this an error in the solution manual or are all our thought processes just wrong?