Photoelectric Effect Post-Module Assessment

Moderators: Chem_Mod, Chem_Admin

gabbym
Posts: 76
Joined: Wed Apr 11, 2018 3:00 am

Photoelectric Effect Post-Module Assessment

Postby gabbym » Wed Oct 17, 2018 12:04 pm

In a second experiment a shorter wavelength light source is used resulting in ejected electrons with a kinetic energy of 4.200 x 10-19 J.
What is the energy of this incident light?
What is the wavelength of this incident light?
The first experiment gives you the work function 3.607 * 10^-19, with a kinetic energy of zero. I understand how to find the energy of the incident light, which is 7.807*10^-19, but I don't know how to find the wavelength. Can someone help me?

Dakota_Campbell_1C
Posts: 51
Joined: Fri Sep 28, 2018 12:15 am

Re: Photoelectric Effect Post-Module Assessment

Postby Dakota_Campbell_1C » Wed Oct 17, 2018 1:03 pm

Once you've found the energy you can use the formula λ=(hc)/E to find wavelength.

Hai-Lin Yeh 1J
Posts: 89
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time

Re: Photoelectric Effect Post-Module Assessment

Postby Hai-Lin Yeh 1J » Wed Oct 17, 2018 4:43 pm

You know the energy which is 7.807 x 10^-19 J. If you want to find the wavelength, use the formula λ=(hc)/E to find the wavelength:

λ=(6.626 x 10^-34 J.s^-1)(3.00 x 10^8 m.s^-1) / (7.807 x 10^-19 J) = 2.54 x 10^-7 m or 254 nm

OR you can take the longer step, which is first finding out the frequency, using E= hv. Then you can plug that into the formula: c= λv. Either way works, it's just the second way is longer whereas the first method is a combination of both equations (E=hv and c=λv).


Return to “Photoelectric Effect”

Who is online

Users browsing this forum: No registered users and 1 guest