Atomic Spectra Module Q 29

H-Atom ()

Moderators: Chem_Mod, Chem_Admin

ClaireHW
Posts: 60
Joined: Fri Sep 29, 2017 7:07 am
Been upvoted: 1 time

Atomic Spectra Module Q 29

Postby ClaireHW » Mon Oct 16, 2017 8:42 pm

In 1.0 s, a 60 W bulb emits 11J of energy in the form of infrared radiation (heat) of wavelength 1850 nm. a) What is the energy per photon of light emitted? b) How many photons are emitted in one second?

I am particularly confused by part b.

(Claire Woolson Dis. 3J)

Mariane Sanchez 1E
Posts: 57
Joined: Fri Sep 29, 2017 7:07 am

Re: Atomic Spectra Module Q 29

Postby Mariane Sanchez 1E » Mon Oct 16, 2017 9:09 pm

Hi!

So basically what I did here is I divided the Total number of Energy released in 1 sec, which is 11J, by the energy per photon, which is the answer you got from part a. After that, you should get the number of photons.

K Stefanescu 2I
Posts: 68
Joined: Fri Sep 29, 2017 7:04 am

Re: Atomic Spectra Module Q 29

Postby K Stefanescu 2I » Mon Oct 16, 2017 9:10 pm

This question is similar to 1.27, for further practice/reference.

a) The words "energy per photon" should bring to mind the formula E=hv. Because we have been given the wavelength instead of frequency, we must rewrite the formula by substituting v=c/w. (I am using w to denote wavelength.) The formula becomes as follows: E=(hc)/w.
Now, plug in the value given and solve for energy.
E=(6.63*10^-34*3*10^8)/(1850*10^-9)= 1.075*10-19 J

b) This question is unusually phrased, in that it has already given you the amount of energy in J that is produced. (More on why this is strange at the end.) But, let's ignore the potentially extraneous information and solve it. If we know that the total amount of energy is 11J and we know the energy per one photon, we can calculate the number of photons emitted as follows by cancelling the units:
11J * ((1 photon)/ (1.075*10^-19J))= 1.0 * 10^20 photons


Given the information, I believe that is what the answer should be. However, keep reading if you want more info on how this question in the module differs from that in the homework.
In question 1.27, we are simply given a 32W bulb, another wavelength value, and asked to solve essentially the same problem: determine the number of photons emitted, this time in 2 seconds instead of 1 second. In order to solve for the total number of photons emitted from the Watt value of the bulb, we would need to do some conversions. 1 W= 1 J/s.
From this, we would be able to determine the amount of energy emitted by the bulb in 2 seconds as follows: 32W = (32J/ 1 s) * 2s = 64J
From there, we would solve the problem as we did the one from the module. So, just be very mindful of what information the question gives you.


Return to “Bohr Frequency Condition, H-Atom , Atomic Spectroscopy”

Who is online

Users browsing this forum: No registered users and 1 guest