## Photoelectric Effect Module

Ashish Verma 2I
Posts: 59
Joined: Fri Sep 28, 2018 12:28 am

### Photoelectric Effect Module

I had a question about one of the parts of the Photoelectric Effect Module Post Assessment.
Question 30 specifically. The given information is as follows:

Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x $10^5$ m/s. The work function for sodium is 150.6 kJ/mol.

The question asks for the frequency of the incident light on the sodium metal surface. So I began by solving for the energy of the incoming light by adding the work function to the energy of the emitted electron (answered in a previous question as 1.99 x $10^-19$ J. Once I had that energy calculation I used Planck's equation to solve for frequency and got 2.27 x $10^38$ Hz. Which was a multiple choice answer that was wrong.

I'm not sure how to properly solve this question and would appreciate any help! Thank you!

Sierra Cheslick 2B
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

### Re: Photoelectric Effect Module

Before adding the work function to the energy of the emitted electron, you must first divide the work function by 6.022 x 10^23, as the work function is given in moles and you are looking for the threshold energy to remove a single electron. You must also then convert this number into joules from kJ so that your units are consistent.

Gabriela Aguilar 4H
Posts: 30
Joined: Fri Sep 28, 2018 12:29 am

### Re: Photoelectric Effect Module

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.
A. What is the kinetic energy of the ejected electron?

How would one go about to solve this problem?
What equation would be used?

Reva Kakaria 1J
Posts: 62
Joined: Fri Sep 28, 2018 12:23 am
Been upvoted: 1 time

### Re: Photoelectric Effect Module

Gabriela Aguilar 4H wrote: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.
A. What is the kinetic energy of the ejected electron?

How would one go about to solve this problem?
What equation would be used?

You would use the equation to calculate the kinetic energy of an object, which is 1/2(mv^2). The velocity of the electron is given, and the mass, which is 9.11 x 10^-31 kg, can be found on the constants and equations sheet.