Early Quantum Question (from Wilberth's Q&A)

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Gabby Magat 3F
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Joined: Wed Sep 30, 2020 10:04 pm

Early Quantum Question (from Wilberth's Q&A)

Postby Gabby Magat 3F » Thu Dec 10, 2020 10:16 am

Hi everyone! I came late to Wilberth's Early Quantum review, so I'm still a little confused on one of the Bohr Frequency Condition problems. Here's the question:

A photon hits an electronically excited hydrogen atom, and its electron goes from the second excited state to the first excited state. What was the energy of the photon?

[Answer Spoiler Warning]
I keep getting 1.6 x 10^-18 J, but the answer is actually 3.0 x 10^-19 J. I feel like I'm missing something really important, but I think I have the right procedure.

Could someone explain how they got the right answer for this problem?

Alexandra Ahlschlager 1L
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Joined: Wed Sep 30, 2020 9:59 pm

Re: Early Quantum Question (from Wilberth's Q&A)

Postby Alexandra Ahlschlager 1L » Thu Dec 10, 2020 10:44 am

What confused me at first was the vocabulary of “first excited state and second excited state” so maybe this is your problem too? The first excited state would be n=2, (not 1 since n=1 is the ground state), and the second excited state would be n=3. So you would use the equation -deltaE = -hR/nf^2 - (-hR/ni^2), where n=3 would be your n-initial and n=2 would be your n-final. (also delta E is negative since this is an emission problem). From there, you should be able to plug in the rest of the values and get the right answer. Hope this helps!

Gabby Magat 3F
Posts: 114
Joined: Wed Sep 30, 2020 10:04 pm

Re: Early Quantum Question (from Wilberth's Q&A)

Postby Gabby Magat 3F » Thu Dec 10, 2020 10:56 am

Alexandra Ahlschlager 1H wrote:What confused me at first was the vocabulary of “first excited state and second excited state” so maybe this is your problem too? The first excited state would be n=2, (not 1 since n=1 is the ground state), and the second excited state would be n=3. So you would use the equation -deltaE = -hR/nf^2 - (-hR/ni^2), where n=3 would be your n-initial and n=2 would be your n-final. (also delta E is negative since this is an emission problem). From there, you should be able to plug in the rest of the values and get the right answer. Hope this helps!


Ohh I see this makes so much more sense, thank you very much!!

Raashi Chaudhari 3B
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Joined: Wed Sep 30, 2020 9:31 pm

Re: Early Quantum Question (from Wilberth's Q&A)

Postby Raashi Chaudhari 3B » Thu Dec 10, 2020 4:12 pm

I was confused on a different question in this review.
The Uncertainty Problem #1 asks, Take the avg diameter of an alveoli, tiny sacs of air in the lungs, to be 0.23mm. If an oxygen molecule is trapped within the sac, the uncertainty in the position will be 0.12mm. What is the minimum uncertainty in its velocity?

The answer is {8.3 x 10^-6 m/s} but I am getting something else.

What should be value of each variable in the equation to solve this?

Brian Bui 3H
Posts: 101
Joined: Wed Sep 30, 2020 9:42 pm

Re: Early Quantum Question (from Wilberth's Q&A)

Postby Brian Bui 3H » Thu Dec 10, 2020 6:36 pm

Raashi Chaudhari 2D wrote:I was confused on a different question in this review.
The Uncertainty Problem #1 asks, Take the avg diameter of an alveoli, tiny sacs of air in the lungs, to be 0.23mm. If an oxygen molecule is trapped within the sac, the uncertainty in the position will be 0.12mm. What is the minimum uncertainty in its velocity?

The answer is {8.3 x 10^-6 m/s} but I am getting something else.

What should be value of each variable in the equation to solve this?


= 0.12 mm = 0.00012 m, so using the equation

You will get >= 4.39 x 10-31 kg m/s

Since , you now have >= 4.39 x 10-31, so you need to divide 4.39 x 10-31 by the mass of one O2 molecule.

1 O2 molecule x (1 mol O2 / 6.022 x 1023 molecules) x (32 g O2 / 1 mol O2) x (1 kg / 1000 g) = 5.3 x 10-26 kg

So divide 4.39 x 10-31 kg m/s by the mass of one O2 molecule (5.3 x 10-26 kg) and you will get 8.3 x 10-6 m/s.

Hope this helps! :)


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