Atomic spectra post assessment question 42

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Eugene Chung 3F
Posts: 142
Joined: Wed Nov 15, 2017 3:03 am

Atomic spectra post assessment question 42

Postby Eugene Chung 3F » Sun Oct 27, 2019 2:05 pm

I tried using this formula first,
1.14 x 10^14 = 3.29x10^15((1/n^2)- (1/16)
But i ended up getting n=3.

So i just tried with n=5 and n=6 and chose 6.
Is the above formula correct? Or am i missing something?
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Jack Riley 4f
Posts: 100
Joined: Sat Aug 24, 2019 12:17 am

Re: Atomic spectra post assessment question 42

Postby Jack Riley 4f » Mon Oct 28, 2019 2:01 pm

The above formula is correct and when I calculated it out I got n=6 as my answer. Maybe you just entered it into your calculator wrong

Sahil Jog 1F
Posts: 126
Joined: Thu Jul 11, 2019 12:16 am

Re: Atomic spectra post assessment question 42

Postby Sahil Jog 1F » Mon Oct 28, 2019 2:03 pm

You were really close, but the formula goes like this. Frequency = R ((1/nf^2) - (1/ni^2)), where nf means n (final) indicating that this is where the electron will be at the end and ni meaning n (initial) indicating where it started. I think you switched your initial and final values, resulting in a different result.


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