Photoelectric Effect PostMod #34

Moderators: Chem_Mod, Chem_Admin

Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

Photoelectric Effect PostMod #34

Postby 005115864 » Wed Oct 10, 2018 2:49 pm

Hi, I need help answering part b of this problem. Here is the entire question;

33. Molybdenum metal must absorb radiation with a minimum frequency of 1.09 x 10^15 s-1 before it can emit an electron from its surface. Answer the following two questions.
A. What is the minimum energy needed to produce this effect

34. B. If molybdenum is irradiated with 194 nm light, what is the maximum possible kinetic energy of the emitted electrons?

For part A, I found the answer was 7.22 *10^-19 because I identified the given 1.09*10^15 as frequency (curly v) so I plugged it into E = Freq * h and got 7.22*10^-19.

Now, I don't know how to approach part B. I know I can use the given information and my answer from Part A to apply it to Part B. So, in part B, I have identified 194nm as my wavelength and know I can use E= (h*c)/wavelength to find my total E but I'm not sure where to go from there.

Anushi Patel 1J
Posts: 61
Joined: Fri Sep 28, 2018 12:19 am
Been upvoted: 1 time

Re: Photoelectric Effect PostMod #34

Postby Anushi Patel 1J » Wed Oct 10, 2018 3:45 pm

Use the wavelength of 194 nm to find the energy of the photons being shined at the metal, which I found to be 1.02x10^-18 J. When the question is asking for the maximum kinetic energy of an emitted electron, I believe it's just asking for the Ek that we have calculated before in class. We know that the energy of the photon-the work function=the kinetic energy of the electrons. Plug in the values into that equation, and subtract the work function you solved for in part A (7.22x10^-19 J) from the total energy of the photon (1.02x10^-18 J), and what's left (3.03x10^-19 J) is the maximum kinetic energy of the emitted electrons. Hopefully that is the right answer, since I don't have the answer key. Hope that helps!

Return to “Properties of Light”

Who is online

Users browsing this forum: No registered users and 3 guests