Test 2 Q4 help  [ENDORSED]

H-Atom ()

Moderators: Chem_Mod, Chem_Admin

Tina Wen 1G
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

Test 2 Q4 help

Postby Tina Wen 1G » Fri Apr 27, 2018 3:52 pm

How should we solve the difference in energy for Q4 in Test 2? And what is the final answer? Thank you

KC Navarro_1H
Posts: 33
Joined: Fri Apr 06, 2018 11:04 am

Re: Test 2 Q4 help  [ENDORSED]

Postby KC Navarro_1H » Fri Apr 27, 2018 10:35 pm

This one confused me at first because how it's written gives off the impression that it uses the E_n = -hR/n^2 equation at some point, but it just uses c = lambda * v and E = hv. Because the red colored flame is a result of the transition, it indicates that the transition has already been done and the measured wavelength of the red colored flame is 700 nm and they just want you to find the energy given off from that transition (sorry if this part is written weird! I don't really know how to explain this well).

To find frequency (v):

c = lambda * v

v = 3.00 x 10^17 nm/s / 700 nm = 4.28 x 10^14 Hz

To find energy (E):

E = hv = (6.626 x 10^-34)(4.28 x 10^14) = 2.83 x 10^-19 J

The difference in energy would be 2.83 x 10^-19 J. I hope this helps!

Gabi Landes 1-H
Posts: 31
Joined: Thu Feb 22, 2018 3:00 am

Re: Test 2 Q4 help

Postby Gabi Landes 1-H » Sun Apr 29, 2018 10:41 am

The way this question was written really through me for a loop. I definitely was looking for a reason to use the Rydberg equation, however the given wavelength was my biggest indication that this was a c=lamda*v & E=hv problem. Using the word "difference" in the question made me think I had to subtract something. That is where I went wrong! Thank you ^^^^^ for explaining!!

Garrett Dahn 1I
Posts: 28
Joined: Fri Apr 06, 2018 11:02 am
Been upvoted: 1 time

Re: Test 2 Q4 help

Postby Garrett Dahn 1I » Sun Apr 29, 2018 11:07 am

I, too, was thrown off by this question, and went through about three separate convoluted attempts before arriving at the far simpler E = hv process. I think what finally convinced me that this was in fact an E = hv problem was the wording of the last two sentences: "a scientist claims that the color arises from the transition of an electron from its excited state to its ground state. [we know this, and we also know that the energy emitted from that transition is going to be a photon) What is the difference in energy between those two states?" That difference in energy, the energy emitted by the transitioning electron, is going to be the energy of the photon, since as an electron moves from a higher n to a lower n, energy in the form of a photon is shed -- during that shedding of energy, we get the emission of light, in this case, red light. And because the question doesn't specify a single n level, we are able to use the E = hv equation to find our answer. Hope this can be useful to someone!

Return to “Bohr Frequency Condition, H-Atom , Atomic Spectroscopy”

Who is online

Users browsing this forum: No registered users and 2 guests