Photoelectric Effect Post-Module Assessment 32B, 33  [ENDORSED]

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Connie2I
Posts: 29
Joined: Fri Sep 29, 2017 7:07 am

Photoelectric Effect Post-Module Assessment 32B, 33

Postby Connie2I » Fri Oct 13, 2017 9:35 pm

32. B. 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?

I used the Energy of a photon = (threshold)(kinetic energy) and got 1.515 * 10^-37 J, 131.1 nm, but it wasn't the right answer, so I was unsure how to solve this

33. Molybdenum metal must absorb radiation with a minimum frequency of 1.09 x 1015 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?
B. If molybdenum is irradiated with 194 nm light, what is the maximum possible kinetic energy of the emitted electrons?

I got the right answer for A, which is 7.22 * 10^-19 J, but I wasn't sure how to solve for kinetic energy given the wavelength.

Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

Re: Photoelectric Effect Post-Module Assessment 32B, 33

Postby Kelly Kiremidjian 1C » Fri Oct 13, 2017 9:50 pm

For 32B you know Kinetic energy of the electron and you know the work function so you are able to solve for the energy of the photon.
KE=4.200 x10^-19 J
Work function=3.607 x 10^-19 J, which you know from the previous question because there was zero kinetic energy.

So then you do:
Energy(photon)=Kinetic Energy(electron) +Work function
Ep=4.200X10^-19J +3.607 X10^-19 J
Ep=7.807 X10-19 J

Then to find wavelength you can use the equation
wavelength=hC/E where h and C are constants and E is the Energy of the photon that you just found, and you should get 255nm.

hope this helped!

Kendall Schemmer 1I
Posts: 20
Joined: Fri Sep 29, 2017 7:03 am

Re: Photoelectric Effect Post-Module Assessment 32B, 33  [ENDORSED]

Postby Kendall Schemmer 1I » Mon Oct 16, 2017 9:37 am

For 33B, you can use the equation Energy(photon)- Energy(remove electron) = E(kinetic).

To find the energy of the photon, use the wavelength given and convert it to meters: 194nm = 194 x 10^-9 m.
Then, using c=ƛv, solve for the frequency of the photon (v), and then use the E=hv equation to solve for the energy of the photon.

The energy you found in part A is the energy required to remove the electron, or the work function.

Subtract the two energy values you found and that will give you the kinetic energy!


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