## 1A.15

H-Atom ($E_{n}=-\frac{hR}{n^{2}}$)

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Chris Tai 1B
Posts: 102
Joined: Sat Aug 24, 2019 12:16 am

### 1A.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.

Which equation(s) would I use to solve this problem?
I'm still a bit confused as to why the ultraviolet spectrum corresponds to the energy level n=1 while the visible light spectrum corresponds to the energy level n=2, could someone help clarify why that's the case?

madawy
Posts: 81
Joined: Fri Aug 09, 2019 12:17 am

### Re: 1A.15

The fact that the emission is ultraviolet tells you that this transition is in the Lyman series, so n2=1
-Convert 102.6 nm into the energy E using the two relations: $E=hv$ and $\lambda \nu = c$

-Then plug into the rydberg formula and solve for n1

Balmer series is the visible region of the light spectrum and Lyman is the UV region. When the electron jumps from quantum level n=2 to n=1, this releases much more energy than when it jumps from n=3 to n=2.
Lyman's principle quantum level is n=1 because UV light has higher energy.
Balmer's principle quantum level is n = 2 because visible light has lower energy.

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