Photoelectron Effect: POST/PRE Module Assessment #28-30


Moderators: Chem_Mod, Chem_Admin

JaylinWangDis1L
Posts: 91
Joined: Wed Sep 30, 2020 10:09 pm

Photoelectron Effect: POST/PRE Module Assessment #28-30

Postby JaylinWangDis1L » Thu Oct 08, 2020 10:10 pm

Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 10^5 m.s^-1. The work function for sodium is 150.6 kJ.mol^-1. Answer the following three questions.
A. What is the kinetic energy of the ejected electron?
B. How much energy is required to remove an electron from one sodium atom?
C. What is the frequency of the incident light on the sodium metal surface?

I watched the video module and I'm still confused. I assume for part A we use E= hV - work function. I'm not sure what to do for part B and for part C I think we use E(photon) = threshold energy + Ek? Can someone break this problem down and explain for me?
(I think I should've posted this in properties of electrons but I have no clue how to edit that)

John Pham 3L
Posts: 141
Joined: Wed Sep 30, 2020 9:49 pm
Been upvoted: 4 times

Re: Photoelectron Effect: POST/PRE Module Assessment #28-30

Postby John Pham 3L » Thu Oct 08, 2020 10:37 pm

For A, you should use the KE equation.
- KE = .
- Just plug in m/s and mass of Electron kg to solve

For B, you are given the amount of energy needed in KJ per mole.
- Convert from KJ to Joules
- Convert moles to atoms using Avogadro's number atoms
- You should end up with Joules per Sodium atom

For C, you need to first calculate the KE from part A and the energy required for one Sodium atom from part B.
- Use E(Photon) = E(Threshold Energy) + E(Kinetic Energy) to solve for Photon energy
- Take Photon's energy and plug into E = hv, solving for frequency

Yun Su Choi 3G
Posts: 90
Joined: Wed Sep 30, 2020 10:09 pm

Re: Photoelectron Effect: POST/PRE Module Assessment #28-30

Postby Yun Su Choi 3G » Fri Oct 09, 2020 9:42 pm

For Part C, I followed through those steps, but I still got the wrong answer. Can someone point out where I went wrong?

Ephoton= 1.506 x 10^5 + 1.99x10^-19 (answer from part a)
Ephoton= 150600

E=hv
V=E/h
V= 150600/ (6.626 x 10^-34)
V= 2.27 x 10^38

John Pham 3L
Posts: 141
Joined: Wed Sep 30, 2020 9:49 pm
Been upvoted: 4 times

Re: Photoelectron Effect: POST/PRE Module Assessment #28-30

Postby John Pham 3L » Sat Oct 10, 2020 8:18 am

Yun Su Choi 2I wrote:For Part C, I followed through those steps, but I still got the wrong answer. Can someone point out where I went wrong?

Ephoton= 1.506 x 10^5 + 1.99x10^-19 (answer from part a)
Ephoton= 150600


You haven't converted from Joule per mole to Joule per atom
1.506*10^5 is still in Joules per mole. Just divide by Avogadro's constant to get Joule per atom.
Then you can plug in v = E / h to solve for frequency

Madison Muggeo 3H
Posts: 95
Joined: Wed Sep 30, 2020 9:35 pm

Re: Photoelectron Effect: POST/PRE Module Assessment #28-30

Postby Madison Muggeo 3H » Sat Oct 10, 2020 3:59 pm

Thank you for this thread! I got stuck on #29 but this really helped!

Tiao Tan 3C
Posts: 100
Joined: Wed Sep 30, 2020 9:59 pm

Re: Photoelectron Effect: POST/PRE Module Assessment #28-30

Postby Tiao Tan 3C » Tue Oct 13, 2020 1:55 am

John Pham 2B wrote:
Yun Su Choi 2I wrote:For Part C, I followed through those steps, but I still got the wrong answer. Can someone point out where I went wrong?

Ephoton= 1.506 x 10^5 + 1.99x10^-19 (answer from part a)
Ephoton= 150600


You haven't converted from Joule per mole to Joule per atom
1.506*10^5 is still in Joules per mole. Just divide by Avogadro's constant to get Joule per atom.
Then you can plug in v = E / h to solve for frequency


Hi. I am a little confused with C too. From which part of the question did you tell that we need to convert to Joule per atom? or is it because question B asked for the energy to remove AN ELECTRON, and question c is a follow-up?

Andrew Jubintoro 3J
Posts: 118
Joined: Wed Sep 30, 2020 9:58 pm
Been upvoted: 1 time

Re: Photoelectron Effect: POST/PRE Module Assessment #28-30

Postby Andrew Jubintoro 3J » Tue Oct 13, 2020 2:11 am

When you use E=hv to solve for the frequency remember that the Energy here is the energy of a single photon. And one photon can only eject one electron (from one atom). Therefore, E(photon) = E(to remove one e-) + E(excess energy in the form of kinetic energy)


Return to “Properties of Light”

Who is online

Users browsing this forum: No registered users and 1 guest