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### photoelectric effect post-module assessment

Posted: Thu Oct 12, 2017 10:29 pm
28. "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.

"how much energy is required to remove an electron from one sodium atom?
A) 2.501 x 10^-22 J
B) 1.506 x 10^5 J
C) 2.501 x 10^-19 J
D) 9.069 x 10^28 J
E) none of the above

This is a problem from the photoelectric effect module. I chose B thinking that the work function which is 150.6 KJ so that's how much energy it would take to remove an electron. I must be overlooking something because B is not the right answer. What am I not noticing?

### Re: photoelectric effect post-module assessment  [ENDORSED]

Posted: Thu Oct 12, 2017 11:00 pm
The question asks for the amount of energy needed to remove an electron from a single sodium atom. In order to solve, you take the work function, convert it to joules, and then divide by Avagadro's number to get the answer.

150.6 KJ/mol is 150600 J/mol. 150600 J/mol divided by 6.02 x 10^23 atoms/mol is 2.501 x 10^-19. The answer would be C.

### Re: photoelectric effect post-module assessment

Posted: Fri Oct 13, 2017 3:58 pm
To clarify, the question is asking about the energy (in J) required to remove one sodium atom. However, they gave you the work function in KJ/mol and you want J/sodium atom which is why you must divide your answer by avogadro's number. Hope this helps!

### Re: photoelectric effect post-module assessment

Posted: Sun Oct 15, 2017 4:58 pm
I have a question pertaining to #30.

What is the frequency of the incident light on the sodium metal surface? I added the work function with the kinetic energy and got the energy of the incident light, but when I divided it by Planck's constant, I got 2.27 * 10^38 (answer B). This answer is incorrect. Can someone catch the err of my ways?

### Re: photoelectric effect post-module assessment

Posted: Sun Oct 15, 2017 4:59 pm
I have a question pertaining to #30.

What is the frequency of the incident light on the sodium metal surface? I added the work function with the kinetic energy and got the energy of the incident light, but when I divided it by Planck's constant, I got 2.27 * 10^38 (answer B). This answer is incorrect. Can someone catch the err of my ways?

### Re: photoelectric effect post-module assessment

Posted: Wed Oct 18, 2017 11:39 am
Jonathan, maybe you need to post the question as well.