Hey guys! I've been having trouble solving this problem. I can get the correct answer (5.4 E-7m) when I use the work function=hv to solve for wavelength, but I didn't use the 1.7x 10^5 m when I solved the problem. How would you recommend solving this?
Q8. You have a metal, Rubidium, which has a work function of 2.3 eV. If the electrons being ejected from the metal have a wavelength of 1.7 x 10⁵ m, what is the wavelength of the incident light?
Work function and incident light
Moderators: Chem_Mod, Chem_Admin
-
- Posts: 56
- Joined: Wed Sep 30, 2020 9:50 pm
Re: Work function and incident light
Also could I get some help on the follow up question?
(b) Calculate the wavelength of 1 atom of Rubidium, if it somehow was propelled and is traveling at 5.32 x 10⁴ m/s.
mm Rubidium = 85.468 g/mol
(b) Calculate the wavelength of 1 atom of Rubidium, if it somehow was propelled and is traveling at 5.32 x 10⁴ m/s.
mm Rubidium = 85.468 g/mol
-
- Posts: 102
- Joined: Wed Sep 30, 2020 9:47 pm
- Been upvoted: 3 times
Re: Work function and incident light
I believe you did the first part right, and as for part b, I think you would simply use the lamba=h/mv equation. Divide Planck's constant h by massxvolume. To get the mass, you would take the mm of rubidium, 85.468g/mol and divide by avogadro's number. Use the velocity that is given. After plugging in and converting to nm you would get 248.4 nm. I hope this is right or sort of helps, I wasn't too sure on it either, so this is only my attempt at solving it.
Re: Work function and incident light
The previous's student strategy for part b sounds right to me!
This is how I would approach the first part of the problem: First covert the 2.3 eV to Joules. Then, find the kinetic energy of the photon by doing .5(mass of electron)(velocity of electron given)^2. Once you have done that you have to add the work function's value in Joules plus the kinetic energy, and that'll get you the energy of the incident photon. Now to find the wavelength of the incident light, you would use the equation wavelength = hc/E
This is how I would approach the first part of the problem: First covert the 2.3 eV to Joules. Then, find the kinetic energy of the photon by doing .5(mass of electron)(velocity of electron given)^2. Once you have done that you have to add the work function's value in Joules plus the kinetic energy, and that'll get you the energy of the incident photon. Now to find the wavelength of the incident light, you would use the equation wavelength = hc/E
Return to “Photoelectric Effect”
Who is online
Users browsing this forum: No registered users and 1 guest