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Topic: Finding delta E and wavelengths  (Read 12311 times)

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Offline MissDee

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Finding delta E and wavelengths
« 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 (nhi  :rarrow: nlo), 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?

Offline tamim83

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Re: Finding delta E and wavelengths
« Reply #1 on: November 29, 2010, 08:30:05 AM »
Each of the wavelengths you are given corresponds to a .  To get :



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



This should allow you to get for each transition.  
« Last Edit: November 29, 2010, 12:13:12 PM by tamim83 »

Offline rabolisk

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Re: Finding delta E and wavelengths
« Reply #2 on: November 29, 2010, 08:40:37 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.

Offline MissDee

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Re: Finding delta E and wavelengths
« Reply #3 on: November 29, 2010, 10:11:33 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?

Offline rabolisk

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Re: Finding delta E and wavelengths
« Reply #4 on: November 29, 2010, 11:07:43 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..
« Last Edit: November 29, 2010, 11:27:55 AM by rabolisk »

Offline Schrödinger

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Re: Finding delta E and wavelengths
« Reply #5 on: November 29, 2010, 11:18:50 AM »
@tamim83 :
Shouldn't the Rydberg equation be :

$$ \dfrac{1}{\lambda} = R_{H} \mathrm{(\dfrac{1}{n_{f} ^2} - \dfrac{1}{n_{i} ^2})} /$$
« Last Edit: November 29, 2010, 11:40:33 AM by Schrödinger »
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Offline tamim83

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Re: Finding delta E and wavelengths
« Reply #6 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  :)

Offline MissDee

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Re: Finding delta E and wavelengths
« Reply #7 on: November 29, 2010, 12:52:13 PM »
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.  ???

Offline sjb

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Re: Finding delta E and wavelengths
« Reply #8 on: November 29, 2010, 04:05:13 PM »
I think you may need to calculate energies first, from the given wavelengths, then subtract?

Offline Schrödinger

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Re: Finding delta E and wavelengths
« Reply #9 on: November 30, 2010, 05:13:45 AM »
@tamim : How did you type those large brackets? Mine are small  :(
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Offline tamim83

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Re: Finding delta E and wavelengths
« Reply #10 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 nhi to nlo.  So you can use them to get the difference in energy between the two levels (one wavelength goes with one ).  You do not have to subtract the given wavelengths themselves.  

You can then use the Rydberg equation to get after some algebra.  Once you have all of the values, you may be able to deduce nhi and nlo for each transition.  
« Last Edit: November 30, 2010, 04:13:28 PM by Borek »

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