## Photoelectric Effect PostMod #30

$c=\lambda v$

005115864
Posts: 64
Joined: Fri Sep 28, 2018 12:15 am

### Photoelectric Effect PostMod #30

This question is related to #28 and #29 but I have gotten stuck on #30. The Question is..

Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1.

What is the frequency of the incident light on the sodium metal surface?

I have an idea of how to approach this problem, however I've gotten stuck. Here's what I'm thinking,

The incident light refers to the Total Light reflected onto the metal surface so this known as the Ep or in an equation (c=lambda*frequency) or (E= h*frequency). However, I don't know lambda or the wavelength and I am looking for the frequency so I cannot use the first equation. But, I do know what H is which Planck's constant and I believe E is equal to (1/2)*Mass of Electron * Velocity of electron^2

but when I try calculating it, I can't seem to get the answer. Can anyone see what I'm doing wrong?

Quinn_Simpson_3D
Posts: 12
Joined: Fri Sep 28, 2018 12:26 am

### Re: Photoelectric Effect PostMod #30

In order to solve this problem, you need to know the mass of an electron, which is 9.11 x 10^-31 kg. You also know the velocity of the electron, which is 6.51 x 10^5 m/s. Since you have both these values, you can plug them into the equation (1/2)mv^2. The value you get is 1.99 x 10^-19 J . Once you have this value, you can use the equation E(photon)- E(work function) = E(excess), where E(excess) is 1.99 x 10^-19 J. You are given the threshold value for a mole of photons, but you need to find the threshold energy for a single photon by diving the threshold number by Avogadro's number (6.022 x 10^23). You find that your new threshold value is 2.5 x 10^-19. Based on the equation earlier, if you add E(excess) and E(work function), you can find the energy of a single photon, which is 4.43 x 10^-19 (2.5 x 10^-19 + 1.99 x 10^-19). Once you finally have the energy of a single photon, you can use the equation E=hv to solve for frequency. h is Planck's constant, or 6.63 x 10^-34, and you already know E (4.43 x 10^-19). Once you solve for v (v=E/h), you have your frequency, which is 6.78 x 10^14.