## Module Question 29

Or Fisher 1I
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Joined: Fri Feb 02, 2018 3:00 am

### Module Question 29

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.
B. How much energy is required to remove an electron from one sodium atom?

A. 2.501 x 10-22 J

B. 1.506 x 105 J

C. 2.501 x 10-19 J

D. 9.069 x 1028 J

E. None of the above

RubyLake1F
Posts: 41
Joined: Fri Apr 06, 2018 11:03 am

### Re: Module Question 29

For this part of the question you actually only need to look at one of the given numbers. The "work function" is equivalent to the amount of energy required to eject an electron from a given material. It has given you the work function of sodium as equalling 150.6 kJ.mol-1, which means that 150.5 kJ of energy is required per mole of sodium. If you divide this number by Avogadro's number (6.022x10^23) you will get energy required per sodium atom. Remember to convert from kJ to Joules (multiply by 10^3). I think your answer should be C.

RubyLake1F
Posts: 41
Joined: Fri Apr 06, 2018 11:03 am

### Re: Module Question 29

Actually now that I have typed out how I answered the question, I am wondering why it isn't necessary to divide the final answer by 11? The reason I am thinking you would need to divide by 11 is because there are 11 electrons per sodium atom, and you are just looking for the energy to eject one electron (not all 11). I got the question right when I did the module by doing what I explained above, so I know that C is the right answer, but can anyone explain why it isn't necessary to divide by 11?

Chem_Mod
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### Re: Module Question 29

150.6 kJ/mol = 150600 J/mol

150600 J/mol * (1 mol/[6.022*1023 atoms]) = 2.501 * 10-19 J/atom

There is no 11 needed in the work.

jessicasam
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### Re: Module Question 29

Does anyone know why we don't add the kinetic energy to the energy to remove e-?

I thought that to find the energy needed to eject the e- we had to add the work function to the kinetic energy.

Charlene D 3I
Posts: 45
Joined: Wed Sep 30, 2020 9:54 pm

### Re: Module Question 29

jessicasam wrote:Does anyone know why we don't add the kinetic energy to the energy to remove e-?

I thought that to find the energy needed to eject the e- we had to add the work function to the kinetic energy.

The work function is equivalent to the required energy per mol needed to eject the e-
Therefore we can just use the provided numbers to solve for energy required per sodium atom.