photoelectric effect post assessment #17: work function

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Ivanna_Tang_3B
Posts: 22
Joined: Wed Sep 21, 2016 2:58 pm

photoelectric effect post assessment #17: work function

Postby Ivanna_Tang_3B » Mon Oct 03, 2016 10:52 pm

Hi, #17 on the photoelectric effect post assessment #17 is as follows:
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.

I was wondering what it meant by the work function and how to use the 150.6 kJ*mol^-1 in Ek=(1/2)mv^2. thanks!

Annie Chang 3G
Posts: 16
Joined: Wed Sep 21, 2016 2:59 pm

Re: photoelectric effect post assessment #17: work function

Postby Annie Chang 3G » Mon Oct 03, 2016 11:26 pm

Hello,
Work function, also known as the threshold energy, is the energy required to remove an electron. So, 150.6 kJ*mol^-1 is needed to eject an electron from the surface of the sodium metal surface.
The work function value does not need to be used in the equation, Ek=(1/2)mv^2.
If you are trying to find the kinetic energy of the ejected electron, you would plug in the mass of the electron for m and the given velocity, 6.61 x 10^5 m*s^-1 for v. The mass of an electron is a constant, 9.10938 x 10^-31kg.

JulissaLopez1F
Posts: 12
Joined: Wed Sep 21, 2016 2:57 pm

Re: photoelectric effect post assessment #17: work function

Postby JulissaLopez1F » Fri Oct 07, 2016 12:29 am

During my discussion section (4L), the class was solving for velocity of E given threshold energy and wavelength of E. For some reason we multiplied the given threshold energy by 1 electron voltage (1.602x10^-19). Why would we need to do that?? Follow up question: what are the units of electron voltage? and threshold energy??
Thanks in advance.


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