## Module Question 27.A

Danielle Sumilang - 1F
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

### Module Question 27.A

Hello! I had a little difficulty answering a question from the post-module assessment.

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 kinetic energy of the ejected electron?

I know that to find the kinetic energy, you use the equation E(photon)-threshold energy. 150.6 kJ.mol-1 is plugged in for the threshold energy but where does the velocity go into the equation? Thank you!

Chem_Mod
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### Re: Module Question 27.A

For Kinetic Energy, Ek = $\frac{1}{2}mv^{2}$

Erin Li 1K
Posts: 30
Joined: Fri Apr 06, 2018 11:03 am

### Re: Module Question 27.A

with the equation 1/2 times mv^2, you use the KJ/mol to covert the mols to kg for the m part in the equation. The m stands for mass, so you need to convert the mols to kg. Then, Kinetic energy is measure in Joules which is already used in the equation, so some units should cancel out.

Emely Reyna 1F
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Joined: Fri Apr 06, 2018 11:04 am

### Re: Module Question 27.A

I think the m in this equation is the mass of the electron

Chem_Mod
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### Re: Module Question 27.A

Since only the kinetic energy is being asked, you should use E = 1/2m*v^2. m is the mass of the object in question (this case electron) and the velocity of that object.