## 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?

1. AbdullahM

$E =\frac{ hc }{ \lambda}$ 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

I got 3.97, which I rounded to 4.0, then I added (x10^-37) is this correct?

3. AbdullahM

It would be 10^-28 not 37

4. anonymous

how? I did it on my calculator and it said -37

5. AbdullahM

$$\large 10^{-34} \times 10^8 = 10^{-34 +8} = 10^{-26}$$ But then after multiplying the numbers we re-adjust to get -27

6. anonymous

umm okay well thank you very much for you're help. i really appreciate it.

7. aaronq

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

okay thank you @aaronq

9. anonymous

the wavelength is 500.0, so would I add (x10^-9m)?

10. whpalmer4

yes, $$500.0 \text{ nm} = 500.0 * 10^{-9}\text{ m}$$

11. anonymous

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

@whpalmer4

13. whpalmer4

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

@whpalmer4 so it would be 239.4 kJ/mol?

15. whpalmer4

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

:)