# Homework Help: Principle Quantum Number Question

1. Oct 7, 2009

### wilson_chem90

1. The problem statement, all variables and given/known data
The hydrogen spectrum contains a blue line with a wavelength of 434 nm. Photons of blue light are emitted when hydrogens electron drops from the fifth energy level to a lover energy level. What is the lower energy level?

2. Relevant equations
En = 13.6 eV/n^2
E = hc/lambda

3. The attempt at a solution
First i'm going to find E.

E = hc/lambda
= (6.63 x 10^-34 J s)(3.00 x 10^8 m/s) / (4.34 x 10^-7 m)
= 4.583 x 10^-19 J

(1.60 x 10^-19 eV/J) x (4.583 x 10^-19 J)
= 2.86 eV

Next i tried 2 different approaches, im not sure either is correct though. I used En = -(13.6eV/n^2) - (-13.6eV/n^2) and i just used En = -(13.6eV/n^2)

1st approach:
En = -(13.6eV/n^2) - (-13.6eV/n^2)

I rearranged this equation to be n^2 = 2(13.6eV)/En5^2 (i'm pretty sure thats incorrect)
so n^2 = 2(13.6 eV)/2.86(25)
n = sqrt(0.38041958)
n = 0.62

2nd approach:
En = -(13.6eV/n^2)
Now i rearranged this equation to be n^2 = En + 13.6eV
= 2.86eV + 13.6eV
n = sqrt16.46 eV
= 4.06

So im pretty sure my problem lies within rearranging the equations. Any suggestions? thanks

2. Oct 7, 2009

### wilson_chem90

I also just realized i can rearrange En = -(13.6eV/n^2) to be n^2 = 13.6eV/En

so using that equation it would be n^2 = 13.6eV/2.86eV
n = sqrt4.75
= 2.2

this one makes alot more sense, seeing how the units cancel

3. Oct 7, 2009

### SmashtheVan

it looks like your problem is with the relationship between the energy of the photon and the change in energy levels.

$$E_{photon}= E_{f}- E_{i}$$

where the initial state is the 5th energy level, E of the photon you figured out was 2.86, and the final state is what we're solving for.

so,
$$E_{n}= E_{photon} + E_{5}$$

and

$$\frac{13.6}{n^{2}} = 2.86 + \frac{13.6}{5^{2}}$$

now its just algebra to find out n

4. Oct 7, 2009

### wilson_chem90

thanks, i thought it was that, but i might've messed up the rearrangement, im not very strong when it comes to that... i got n^2 = 13.6eV - 0.544eV/2.86eV (the 0.544 is 13.6/5^2)
I worked it out and got an answer of 2.14

5. Oct 7, 2009

### SmashtheVan

i still think you're doing something a bit off here. the .544 and 2.86 should be summed, not divided. and then divide 13.6 by the new sum for your answer. it should give you n^2=(approx.)4, and n=2

remember that n is a discrete energy level, which are integers, and if your decimal is off by more than a tenth or two from the whole, something might be awry in your calculations.

Last edited: Oct 7, 2009
6. Oct 7, 2009

### wilson_chem90

but when i do that i get a decimal, not a whole number. and its 0.544