1A 15


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Jamie Lee 1H
Posts: 54
Joined: Fri Aug 09, 2019 12:15 am

1A 15

Postby Jamie Lee 1H » Sat Oct 12, 2019 10:07 pm

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 specific equation are we supposed to use?

KaitlynBali_4B
Posts: 51
Joined: Fri Aug 02, 2019 12:16 am

Re: 1A 15

Postby KaitlynBali_4B » Sat Oct 12, 2019 10:24 pm

First you would solve for the frequency by rearranging and using this formula: c = λv. Then you would use the Rydberg equation (it is on page 7 of our book).

Here is a link to the Rydberg Equation too:

https://calistry.org/calculate/rydbergEquation

(I had a difficult time typing the equation here).

Ruby Tang 2J
Posts: 102
Joined: Fri Aug 30, 2019 12:15 am

Re: 1A 15

Postby Ruby Tang 2J » Sat Oct 12, 2019 10:28 pm

The Lyman Series is defined as an electron transition from any n greater than or equal to 2 to n = 1 (or vice versa). You would want to use the Rydberg equation in this case: E = -(hR)/n^2. You can substitute E for hv, and then the h's on both sides of the equation cancel out and you're left with v=-R/n^2. You're given the wavelength in the question already, so you can use v=c/λ (rearranged from c= λv) to calculate frequency. Once you have the frequency, you can now calculate the initial and final energy levels. You know one of the energy levels is n=1 already (by definition of the Lyman series, so by rearranging the equation you can get v = R(1/n1^2 - 1/n2^2), and you can plug in v, R, and n1 to solve for n2. Hope this helps!


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