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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?
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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AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.1\[E =\frac{ hc }{ \lambda}\] 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. :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 3.97, which I rounded to 4.0, then I added (x10^37) is this correct?

AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.1It would be 10^28 not 37

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how? I did it on my calculator and it said 37

AbdullahM
 one year ago
Best ResponseYou've already chosen the best response.1\(\large 10^{34} \times 10^8 = 10^{34 +8} = 10^{26}\) But then after multiplying the numbers we readjust to get 27

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0umm okay well thank you very much for you're help. i really appreciate it.

aaronq
 one year ago
Best ResponseYou've already chosen the best response.0After 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 \(kJ/mol\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay thank you @aaronq

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the wavelength is 500.0, so would I add (x10^9m)?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0yes, \(500.0 \text{ nm} = 500.0 * 10^{9}\text{ m}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay 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?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0your 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...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@whpalmer4 so it would be 239.4 kJ/mol?

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0I 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)
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