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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: A5HLEY on October 24, 2007, 09:14:17 PM

Title: Energy of Photons
Post by: A5HLEY on October 24, 2007, 09:14:17 PM
Ok guys, one more question.

An argon ion laser emits light at 489 nm. What is the energy of photons of this wavelength, expressed on a molar basis?

So I know that E=hv, so I calculated E = (6.63E-34)(489)
= 3.24207E-31

but it asks for the answer in kJ/mol, so I'm not sure I have the right units. Any input?
Title: Re: Energy of Photons
Post by: enahs on October 24, 2007, 09:20:41 PM
No, you do not have the right units. Or much else. :(

E=hv
v is frequency, not wavelength. You must convert wavelength to frequency.

Unlike in the problem earlier, this time you do have to use both equations.


Title: Re: Energy of Photons
Post by: Sev on October 24, 2007, 09:23:31 PM
Also, don't forget nano = 10-9.
Title: Re: Energy of Photons
Post by: ARGOS++ on October 24, 2007, 09:25:46 PM
Dear A5HLEY,

If I remember correctly for such a short time period, then I have already answered your Question.
Please read the answers you got and with a half of a line you know the correct answer.
The Energy  is = 1.00 “Einstein” of  489nm  - Correct?

Why not using the Diagram as a Hint?



[Edited:]
  For the Diagram and for the "Einstein"'s you may re-read your
  other Question:  "Frequency of Radiation (http://www.chemicalforums.com/index.php?topic=20121.0)”


Good Luck!

                    ARGOS++
Title: Re: Energy of Photons
Post by: A5HLEY on October 24, 2007, 09:31:00 PM
Ok, so let me try again.

So c=(lambda)(v), thus v=c/(lambda)

v = (3E8)/489 = 6.1346.9325 (frequency)

E=hv
= (6.63E-34)(613496.9325)
= 4.0675E-28

I don't know... I'm sorry if its not right guys. :\ I'm thinking I might go discuss this with my teacher tomorrow cause I'm just not getting this like I should be. I understand the basic equations, but I have trouble with the units, etc. Thank you for your help though. :)
Title: Re: Energy of Photons
Post by: Sev on October 24, 2007, 09:35:18 PM
Quote
So c=(lambda)(v), thus v=c/(lambda)

v = (3E8)/489 = 6.1346.9325 (frequency)

E=hv
= (6.63E-34)(613496.9325)
= 4.0675E-28

See my last post.
Title: Re: Energy of Photons
Post by: ARGOS++ on October 24, 2007, 09:41:44 PM

Dear A5HLEY,

NO! it’s much easer as I told :
ΔE = 2.86 * 104 / λ  [kcal/mole] = 119’719 / λ [kJoule/mole]   - if λ in [nm]
   

Remember:
1.00 * Einstein(489)  = 119’719 / λ  =  ??? [kJoule/mole]
(Just a simple Division.)


Good Luck!

                    ARGOS++
Title: Re: Energy of Photons
Post by: Sev on October 24, 2007, 09:56:39 PM
Argos++, you are likely confusing her with this 'Einstein' business.
She should use the method that she has been shown in class.
Title: Re: Energy of Photons
Post by: ARGOS++ on October 24, 2007, 10:19:26 PM

Dear Sev,

SORRY!  - I use “exactly" the same Formula, if you would read.

But you all forgot the Calculation for the unit [mole] and that is the “Avogdro Constant” with:

NA = 6.0221 * 1023 mole-1

If you insert it into the equation you will get the same result and the same Formula!
(1.0 mole Photon is called 1.0 "Einstein", it is not my Invention!!)

Sorry – it’s Mother Nature.


Good Luck!

                    ARGOS++
Title: Re: Energy of Photons
Post by: enahs on October 24, 2007, 10:48:53 PM
Yeah, I have never heard of the Einstein my self. I can do this calculation in my sleep and what you are saying confuses me! heh


to A5hley, your last method was correct, but as you said, units!
Your wavelength was in nM but c is in m, and so is h. You must first convert nm to m.

That calculation is for just 1 photon, you need to then find it out for a mol of photons (Multiplying Avagrados number)