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?

nicolely2F
Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: Atomic spectra post assessment question 42

Postby nicolely2F » Mon Oct 28, 2019 11:45 pm

You used the right formula; you just mixed up the place of the n. The formula is v = R (1/n1^2 - 1/n2^2), where n1 = 16. Do it this way and you should get n2 = approx. 5.99 = 6

Ariel Davydov 1C
Posts: 110
Joined: Thu Jul 11, 2019 12:16 am
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Re: Atomic spectra post assessment question 42

Postby Ariel Davydov 1C » Tue Oct 29, 2019 8:35 am

It helps if instead of using the frequency equation you derive it through this basic equation: deltaE = Efinal - Einitial. From here, you can sub in En = -hr(1/n^2) for the final and initial energy values and solve from there. This will help prevent any future mixups with n values since they are assigned a designation of initial or final.


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