# Frequency question

1. Oct 12, 2006

### lorka150

For electromagnetic radiation with a wavelength of 576.3nm, what is the frequency of the radiation in s^-1? What is the energy in J of one photon of the radiation? What is the energy in kJ of one mole of photons of the radiation?

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I think I have the answer.

For the frequency, I did nu=c/lamba

nu = 2.9979x10^8/(576.3)
nu = 520197.8136
= 5.202 x 10^5

For the energy in J, I did:
6.6256x10^-34 x 5.202x10^5 = 3.447 x 10^-28

For the last part, I figured out the mols but don't know where to go from there:

3.447x10^-28 x 6.022x10^23 = 0.0002076mols

2. Oct 12, 2006

### lkh1986

Yeap, everything seems okay for me. Just have to be careful with the SI unit, for example, convert pm to To calculate the energy in kJ of one mole of photons of the radiation, multiply the answer obtained in part (b) with Avogadro constant. Then you will get the answer in unit Joule. remember to divide it by 1000 to change the unit to the required KJ.

3. Oct 16, 2006

### lorka150

lkh - i am not sure what you mean - is that for a)?

4. Oct 16, 2006

### Astronuc

Staff Emeritus
lkh1986 mentioned several things.

1. Careful with units. If c is given in m/s, then wavelength must be in m, in order to give frequency in s-1. 1 nm (nanometer) = 10-9m

2. E = h$\nu$, where E is energy, h is Planck's constant and $\nu$ is frequency. Again make sure units are consistent.

Here one is presumably being asked for the energy of 6.0223 x 1023 (Avogadro's number) photons of the energy calculated previously for a single photon. 1 kJ = 1000 J.

5. Oct 16, 2006

### lorka150

astronuc, thanks.

so for
the first one, instead of using 576.3nm, I would

then multiply that by 10^-9m/1nm, so I get 5.76 x 10-7?

6. Oct 16, 2006

### Astronuc

Staff Emeritus
Correct. It is also good practice, to write the units with the number, e.g.
576.3nm = 5.76 x 10-7 m, so that one is sure that the correct units are being used.

There have been some classic failures, such as sending a multimillion dollar satellite crashing into Mars rather than orbit because two teams used different units when data was transferred from one to the other.

7. Oct 16, 2006

### lorka150

Thank you for your help with this.

So I divided 2.9979x10^8 m/s by 5.76 x 10^-7 m
to get nu= 5.20 x 10 ^14 s^-1

Then for the second part (what is the energy in J of one photon of the raditon, I would still multiply 5.20 x 10 ^14 s^-1 by 6.6256 x 10 ^-34 J.sec, and then I would get 3.45 x 10 ^-19 J.

For the third part (what is the energy of one mole of photons of the raditation in kJ.... ) I did 3.45 x 10^-19 J x 6.022 x 10 ^23 = 207759J, then divided it by 1000 to get kJ, which was 208. kJ.

I hope that's right now! Thanks so far!