question 42 on post assessment

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005199302
Posts: 108
Joined: Fri Sep 28, 2018 12:15 am

question 42 on post assessment

Postby 005199302 » Mon Oct 22, 2018 4:42 pm

An excited hydrogen atom emits light with a frequency of 1.14 x 10^14 Hz to reach the energy level n = 4. In what principle quantum level did the electron begin?

I keep solving and getting the incorrect answer. Can someone please solve this (not using Rydberg equation) and actually compute the numbers? Thanks!

Hai-Lin Yeh 1J
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Joined: Fri Sep 28, 2018 12:16 am
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Re: question 42 on post assessment

Postby Hai-Lin Yeh 1J » Mon Oct 22, 2018 5:00 pm

In the atomic spectra module (video portion), at one point (at 26.04), Professor Lavelle introduces a formula that he derives, which involves the frequency and 2 energy levels. The formula is:

V = R [(1/n1^2) - (1/n2^2)].

So, you know that the frequency (v) = 1.14 x 10^14 Hz. The energy level that it ends up at is 4, so n1 = 4.

Plug into the equation:
1.14 x 10^14 Hz = 3.28x10^15 [(1/4^2) - (1/n2^2)].
Divide both sides by 3.28x10^15.
0.0348 = [(1/16) - (1/n2^2)]
Subtract 1/16 on both sides
-0.0277 = -1/n2^2
Divide by -1.
0.0277 = 1/n2^2
Multiply by n2^2 to get it by itself:
0.0277n2^2 = 1
Divide by 0.0277
n2^2 = 1/0.0277
Square root (1/0.0277):
You get n = 6

005199302
Posts: 108
Joined: Fri Sep 28, 2018 12:15 am

Re: question 42 on post assessment

Postby 005199302 » Mon Oct 22, 2018 5:43 pm

I really appreciate the response and understand how you came to your answer. What's bothering me, however, is that I end up with n^2 being 10.3.
This is my work:

change of e = hv = 7.56 x 10^-20 = Rh/(n1)^2 - Rh/(n2)^2

When I plug n2=4 into the equation and add that to hv that leaves me with the following:
2.18x10^-18/(n1)^2 = 2.12 x 10^-19

Solving for (n1)^2 gives you 10.3

Does anyone know this is wrong?


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