The photoelectric effect is a concern in space, where parts of a spacecraft exposed to the sunlight will develop a net positive charge, which can interfere with or damage sensitive cir The panels that cover the exterior of the Space Shuttle are composed of graphite fibers, which have a threshold frequency of 1.2 x10^15 Hz.

Question

A. What is the K.E of an electron ejected from a panel, in J, when a photon having a wavelength of 190 mm is absorbed?

For this part, I tried using Ek= hc/ lambda and got 1.045x10^-18 J. But the answer key is 2.5x10^-19 J.

B. What is the velocity of ejected electron in m/s?

I know I have to use Ek=1/2 mv^2. Correct me if I'm wrong.

C. What is the de Broglie wavelength of this ejected electron in m?

De Broglie wavelength is lambda= h/mv

Answer 1

a. find the energy of the 190nm photon, you have the right formula. Now subtract the threshold energy (E=h f). With the remainder, you have ke.

b. correct, use E=1/2 m v^2

c. correct formula...

Answer 2

A.
K.E. = hc/w - hf

B
Yes.

C.
Yes.

Answer 3

Thank you so much!