Module question 30 c

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melissa_dis4K
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Joined: Fri Sep 28, 2018 12:28 am

Module question 30 c

Postby melissa_dis4K » Sun Oct 21, 2018 11:48 pm

The question asks "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. What is the frequency of the incident light on the sodium metal surface?" I began by doing E=h*v(frequency). However, I don't know E. So is there something I'm missing to find E first in order to find the frequency?

Heesu_Kim_1F
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am

Re: Module question 30 c

Postby Heesu_Kim_1F » Mon Oct 22, 2018 1:38 am

Problem 30C is a problem that uses the answers that you get for 28A and 29B. You can get the value of E by using the photoelectric equation: E(hv)=work function+Ek (kinetic energy). The kinetic energy value is the answer for 28A and the work function energy value is the answer for 29B. Thus, in order to find the value of E, add those two answers. Then, doing v=E/h, you can find the value of the frequency. Hope this helps!

melissa_dis4K
Posts: 106
Joined: Fri Sep 28, 2018 12:28 am

Re: Module question 30 c

Postby melissa_dis4K » Tue Oct 23, 2018 1:05 pm

Thank you, it did help! I was overthinking the equations. But, now I got 150.6 J for question 29b when I added the work function plus my answer from 28a. I then used frequency=E/h and got 2.27 x 10^35. However, the answer seems to be to the power of 38. Do you know why I get the wrong power sometimes?

melissa_dis4K
Posts: 106
Joined: Fri Sep 28, 2018 12:28 am

Re: Module question 30 c

Postby melissa_dis4K » Tue Oct 23, 2018 1:51 pm

I realized my answer is wrong can someone explain what I did wrong? Thanks!

Julia Lindner 1I
Posts: 35
Joined: Fri Sep 28, 2018 12:17 am

Re: Module question 30 c

Postby Julia Lindner 1I » Wed Oct 24, 2018 3:45 pm

melissa_dis4F wrote:Thank you, it did help! I was overthinking the equations. But, now I got 150.6 J for question 29b when I added the work function plus my answer from 28a. I then used frequency=E/h and got 2.27 x 10^35. However, the answer seems to be to the power of 38. Do you know why I get the wrong power sometimes?


Your answer for 29b should actually be 2.501 x 10^-19. You get this by dividing 1.506 J (work function for 1 mol of sodium) by Avogadro's number to get the work function for individual sodium atoms. You can then add your answers from a and b (1.99 x 10^-19 and 2.501 x 10^-19) to get the energy of the light, and from there you can use v=E/h to get frequency.


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