photoelectric effect

$c=\lambda v$

Katherine Wu 1H
Posts: 104
Joined: Fri Aug 30, 2019 12:15 am
Been upvoted: 2 times

photoelectric effect

Which one of following is not describing the photoelectric effect?
A. E (photon) – E (remove e-) = E (excess)

B. E (photon) – E (remove e-) = EK (e-)

C. hv - work function = 1/2mv2

D. λv = c

E. None of the above
I don't understand which of these are not describing the effect and why.

Rohit Ghosh 4F
Posts: 99
Joined: Thu Jul 25, 2019 12:17 am

Re: photoelectric effect

I think D does not describe the photoelectric effect because it instead just describes light going through a medium.

Haley Pham 4I
Posts: 51
Joined: Fri Aug 09, 2019 12:16 am

Re: photoelectric effect

D. λv = c does not describe the photoelectric effect because the photoelectric experiment was designed to measure the how much energy is required to remove an electron from a metal surface and the kinetic energy of the electron, which is described by answers A, B, and C. Answer choice D describes the wavelength and frequency of light.

DTingey_1C
Posts: 55
Joined: Fri Aug 30, 2019 12:16 am

Re: photoelectric effect

D. λv = c is the correct answer because it doesn't have to do with the photoelectric effect equation or any of its derivations. The equation shows the relationship between a wave's frequency and wavelength.