## 1.A.15

$E=hv$

Danielle DIS 3K
Posts: 35
Joined: Wed Sep 30, 2020 9:38 pm

### 1.A.15

Can someone please explain their step by step process in solving this question?

"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".

Sophia Hu 1C
Posts: 51
Joined: Wed Sep 30, 2020 10:09 pm
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### Re: 1.A.15

Since it states the spectrum is the ultraviolet spectrum of atomic hydrogen, then n_final = 1. The reason for this is because ultraviolet light corresponds to the Lyman series.

delta E = hc/lambda (you can get these equations from E = hv, c = vlambda)
delta E = (6.626 x 10^-34) (3.00 x 10^8) / (102.6 x 10^-9) = 1.94... x 10^-18 J
Remember that this delta E represents the energy of the photon emitted. Therefore, the energy of the electron is - 1.94... x 10^-18 J

En = - hR/n^2
E_final = - (6.626 x 10^-34) (3.29 x 10^15) / (1)^2 = - 2.18... x 10^-18 J

delta E = E_final - E_initial
(- 1.94... x 10^-18) = - 2.18... x 10^-18 - (-hR/n^2)
2.43... x 10^-19 = hR/n^2
2.43... x 10^-19 = (6.626 x 10^-34) (3.29 x 10^15)/n^2
0.111... = 1/n^2
8.99... = n^2
n = 3

Therefore, the transition would be from n = 3 to n = 1

Praneetha Kakarla 1E
Posts: 38
Joined: Wed Sep 30, 2020 9:33 pm
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### Re: 1.A.15

Just to add on to what Sophia said, you can also use the Rydberg equation if you wanted to solve the problem a bit quicker on a test/exam. You would use n1 = 1, find the frequency through c=lambda*nu using the given wavelength, and solve for n2 in the Rydberg equation.