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anonymous

  • one year ago

11. A certain source emits radiation of wavelength 500.0 nm. What is the energy, in kJ, of one mole of photons of this radiation?

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  1. AbdullahM
    • one year ago
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    \[E =\frac{ hc }{ \lambda}\] E = energy of the photon h = Plank's constant = \(\large 6.626 \times 10^{-34}~joule\cdot s\) C = speed of light = \(\large 2.998 \times 10^8 ~m/s\) \(\large\lambda\) = wavelength of photon So, plug in the values into the formula and tell me what you get. :)

  2. anonymous
    • one year ago
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    I got 3.97, which I rounded to 4.0, then I added (x10^-37) is this correct?

  3. AbdullahM
    • one year ago
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    It would be 10^-28 not 37

  4. anonymous
    • one year ago
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    how? I did it on my calculator and it said -37

  5. AbdullahM
    • one year ago
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    \(\large 10^{-34} \times 10^8 = 10^{-34 +8} = 10^{-26}\) But then after multiplying the numbers we re-adjust to get -27

  6. anonymous
    • one year ago
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    umm okay well thank you very much for you're help. i really appreciate it.

  7. aaronq
    • one year ago
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    After that you need to multiply the energy value by Avogadro's constant to get a value in \( J/per ~mole\), then divide by 1000 to get \(kJ/mol\).

  8. anonymous
    • one year ago
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    okay thank you @aaronq

  9. anonymous
    • one year ago
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    the wavelength is 500.0, so would I add (x10^-9m)?

  10. whpalmer4
    • one year ago
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    yes, \(500.0 \text{ nm} = 500.0 * 10^{-9}\text{ m}\)

  11. anonymous
    • one year ago
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    okay so, I have E= hc/wavelength h= 6.626*10^-34 J.S c= 2.998*10^8 m/s E= (6.626*10^-34 J.S)(2.998*10^8 m/s)/500.0*10^-9 m E= 3.97*10^-37 --------------------------------------------------------- **aaron said: Avogadro's constant= 6.02214086*10^23 mol^-1 (to find J/per mole) then /1000 (to find kJ/mol)** --------------------------------------------------------- 3.97*10^-37/6.02214086*10^23 = 6.59*10^-15 6.59*10^-15/1000 = 6.59*10^-18 so the answer would be : 6.59*10^-18, right?

  12. anonymous
    • one year ago
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    @whpalmer4

  13. whpalmer4
    • one year ago
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    your energy figure in the first step is incorrect. \[E = \frac{h c}{\lambda} = \frac{(6.626*10^{-34}\text{ J s})(2.99792458*10^8\text{ m/s})}{500*10^{-9}\text{ m}} \] just looking at the exponents your figure should be \(*10^{-34+8-(-9) =-17}\) times the fractional part, which is roughly \(20/500 \approx 0.04\) so neighborhood of \(4*10^{-19} \text{ J}\) per photon. Not bad for an estimate, I get \(3.97285*10^{-19}\) when I punch it all in... Now, the Avogadro number is \(6.022*10^{23}\) whatevers/mol, so our final answer should be about \[6*4*10^{-19+23}*10^{-3} = 24*10^{1} = 2.4*10^2 \ \ \ \ (\text{kJ/mol})\] You can work out the exact value...

  14. anonymous
    • one year ago
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    @whpalmer4 so it would be 239.4 kJ/mol?

  15. whpalmer4
    • one year ago
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    I got 239.2 kJ/mol, the difference may stem from my value for the speed of light (if I use 3*10^8 I get 239.4)

  16. AbdullahM
    • one year ago
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    :)

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