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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: MissDee on November 28, 2010, 10:47:35 AM

Title: Finding delta E and wavelengths
Post by: MissDee on November 28, 2010, 10:47:35 AM: Hi,

I'm having a little trouble finding :delta: E and wavelengths.

I need to find :delta: E, probable transition (n_hi :rarrow: n_lo), and wavelength calculated (in nm).

Wavelengths Observed:
389.02
397.12
954.86
1005.2

To find :delta: E, I know I have to subtract that higher wavelength from the lower wavelength.

For example, would this be:
(389.02) - (1005.2) = -616.18
(389.02) - (954.86) = -565.84
(389.02) - (397.12) = -8.1

(397.12) - (1005.2) = -608.08
(397.12) - (954.86) = -557.74

(954.86) - (1005.2) = -50.34

For wavelength calculated, I would then:
(1.19627 x 10⁵) / ( :delta: E)

Which would be:
(1.19627 x 10⁵) / (-616.18) = -194.14
(1.19627 x 10⁵) / (-565.84) = -211.41
(1.19627 x 10⁵) / (-8.1) = -14768.76** (is this number suppose to be like that?)

(1.19627 x 10⁵) / (-608.08) = -196.73
(1.19627 x 10⁵) / (-557.74) = -351.9

(1.19627 x 10⁵) / (-50.34) = -2376.38** (is this number suppose to be like that?)

^^ I believe I did these above calculations correctly. Please let me know if I did anything wrong.

Then, to find probable transition, I really don't know what to do. Could someone please help me out here?
Title: Re: Finding delta E and wavelengths
Post by: tamim83 on November 29, 2010, 08:30:05 AM: Each of the wavelengths you are given corresponds to a (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+E+%7D&hash=6263178bf553da79c7994516b335d5dcb96f4a7b). To get (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+E+%7D&hash=6263178bf553da79c7994516b335d5dcb96f4a7b):

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+E+%3D+%5Cfrac%7Bhc%7D%7B%5Clambda%7D+%7D&hash=31252b5a91c6051340e28ea59efbac4092f7a593)

To get the transitions, you need to use the Rydberg equation (if this is for the H atom, if not R_H will be different).

(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5Cfrac%7B1%7D%7B%5Clambda%7D+%3D+R_%7BH%7D+%5Cleft%28+%5Cfrac%7B1%7D%7Bn_%7Bf%7D%5E2%7D+-+%5Cfrac%7B1%7D%7Bn_%7Bi%7D%5E2%7D+%5Cright%29+%7D&hash=ca4c1454b429f90c1a9f852efcf4a103d9bfd3dd)

This should allow you to get (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+n+%7D&hash=a991c22060b591abe06e2fec336930a06fb93064) for each transition.
Title: Re: Finding delta E and wavelengths
Post by: rabolisk on November 29, 2010, 08:40:37 AM: Quote from: MissDee on November 28, 2010, 10:47:35 AM
To find :delta: E, I know I have to subtract that higher wavelength from the lower wavelength.

For example, would this be:
(389.02) - (1005.2) = -616.18
(389.02) - (954.86) = -565.84
(389.02) - (397.12) = -8.1

(397.12) - (1005.2) = -608.08
(397.12) - (954.86) = -557.74

(954.86) - (1005.2) = -50.34

Look at the formula again... You did the algebra wrong.
Title: Re: Finding delta E and wavelengths
Post by: MissDee on November 29, 2010, 10:11:33 AM: Quote from: rabolisk on November 29, 2010, 08:40:37 AM
Quote from: MissDee on November 28, 2010, 10:47:35 AM
To find :delta: E, I know I have to subtract that higher wavelength from the lower wavelength.

For example, would this be:
(389.02) - (1005.2) = -616.18
(389.02) - (954.86) = -565.84
(389.02) - (397.12) = -8.1

(397.12) - (1005.2) = -608.08
(397.12) - (954.86) = -557.74

(954.86) - (1005.2) = -50.34

Look at the formula again... You did the algebra wrong.

Did I switch them by accident? Would it be:

(1005.2) - (389.02) = 616.18
(954.86) - (389.02) = 565.84
(397.12) - (389.02) = 8.1

(1005.2) - (397.12) = 608.08
(954.86) - (397.12) = 557.74

(1005.2) - (954.86) = 50.34

Is everything else correctly, only I plug in these numbers instead of the previous numbers?
Title: Re: Finding delta E and wavelengths
Post by: rabolisk on November 29, 2010, 11:07:43 AM: Quote from: MissDee on November 29, 2010, 10:11:33 AM
Quote from: rabolisk on November 29, 2010, 08:40:37 AM
Quote from: MissDee on November 28, 2010, 10:47:35 AM
To find :delta: E, I know I have to subtract that higher wavelength from the lower wavelength.

For example, would this be:
(389.02) - (1005.2) = -616.18
(389.02) - (954.86) = -565.84
(389.02) - (397.12) = -8.1

(397.12) - (1005.2) = -608.08
(397.12) - (954.86) = -557.74

(954.86) - (1005.2) = -50.34

Look at the formula again... You did the algebra wrong.

Did I switch them by accident? Would it be:

(1005.2) - (389.02) = 616.18
(954.86) - (389.02) = 565.84
(397.12) - (389.02) = 8.1

(1005.2) - (397.12) = 608.08
(954.86) - (397.12) = 557.74

(1005.2) - (954.86) = 50.34

Is everything else correctly, only I plug in these numbers instead of the previous numbers?

Make sure you understand exactly what :delta: E and wavelength refer to..
Title: Re: Finding delta E and wavelengths
Post by: Schrödinger on November 29, 2010, 11:18:50 AM: @tamim83 :
Shouldn't the Rydberg equation be :

$$ \dfrac{1}{\lambda} = R_{H} \mathrm{(\dfrac{1}{n_{f} ^2} - \dfrac{1}{n_{i} ^2})} /$$
Title: Re: Finding delta E and wavelengths
Post by: tamim83 on November 29, 2010, 12:14:59 PM: @Schroedinger

Yeah, you're right. I forgot to add the powers when doing the latex (I intended to do them last). Its fixed now. Thanks :)
Title: Re: Finding delta E and wavelengths
Post by: MissDee on November 29, 2010, 12:52:13 PM: Quote from: rabolisk on November 29, 2010, 11:07:43 AM
Quote from: MissDee on November 29, 2010, 10:11:33 AM
Quote from: rabolisk on November 29, 2010, 08:40:37 AM
Quote from: MissDee on November 28, 2010, 10:47:35 AM
To find :delta: E, I know I have to subtract that higher wavelength from the lower wavelength.

For example, would this be:
(389.02) - (1005.2) = -616.18
(389.02) - (954.86) = -565.84
(389.02) - (397.12) = -8.1

(397.12) - (1005.2) = -608.08
(397.12) - (954.86) = -557.74

(954.86) - (1005.2) = -50.34

Look at the formula again... You did the algebra wrong.

Did I switch them by accident? Would it be:

(1005.2) - (389.02) = 616.18
(954.86) - (389.02) = 565.84
(397.12) - (389.02) = 8.1

(1005.2) - (397.12) = 608.08
(954.86) - (397.12) = 557.74

(1005.2) - (954.86) = 50.34

Is everything else correctly, only I plug in these numbers instead of the previous numbers?

Make sure you understand exactly what :delta: E and wavelength refer to..

I am still really confused. ???
Title: Re: Finding delta E and wavelengths
Post by: sjb on November 29, 2010, 04:05:13 PM: I think you may need to calculate energies first, from the given wavelengths, then subtract?
Title: Re: Finding delta E and wavelengths
Post by: Schrödinger on November 30, 2010, 05:13:45 AM: @tamim : How did you type those large brackets? Mine are small :(
Title: Re: Finding delta E and wavelengths
Post by: tamim83 on November 30, 2010, 08:17:27 AM: @Schroedinger: use
Code: [Select]
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@MissDee: The given wavelengths correspond to the photons emitted during the transition from n_hi to n_lo. So you can use them to get the difference in energy between the two levels (one wavelength goes with one (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+E+%7D&hash=6263178bf553da79c7994516b335d5dcb96f4a7b) ). You do not have to subtract the given wavelengths themselves.

You can then use the Rydberg equation to get (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+n%7D&hash=d3f48a10544dd398365d592f9e74567194ecba05) after some algebra. Once you have all of the (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.forkosh.dreamhost.com%2Fmimetex.cgi%3F%7B+%5CDelta+n+%7D&hash=a991c22060b591abe06e2fec336930a06fb93064) values, you may be able to deduce n_hi and n_lo for each transition.