Photoelectric Effect

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VanessaZhu2L
Posts: 86
Joined: Wed Sep 30, 2020 9:44 pm

Photoelectric Effect

Postby VanessaZhu2L » Thu Oct 15, 2020 5:42 pm

Hi! Can someone explain this?
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.
Answer the following three questions.
A. What is the kinetic energy of the ejected electron?
B. How much energy is required to remove an electron from one sodium atom?
C. What is the frequency of the incident light on the sodium metal surface?

Halle Villalobos 3E
Posts: 85
Joined: Wed Sep 30, 2020 9:52 pm

Re: Photoelectric Effect

Postby Halle Villalobos 3E » Thu Oct 15, 2020 6:00 pm

For part A, use Ek =(1/2)mv^2. The mass for an electron is 9.11*10^(-31) kg.
For part B, you would divide the work function by Avogadro's constant since the units for work function are (kJ/mol)
For part C, you would divide the sum of the kinetic energy and workfunction by Planck's constant

Marisa Gaitan 2D
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Joined: Wed Sep 30, 2020 9:47 pm
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Re: Photoelectric Effect

Postby Marisa Gaitan 2D » Thu Oct 15, 2020 6:03 pm

For A you need to use Ek = 1/2 mv^2 where m is mass of an electron and v is velocity. Then for B you need to take the work function and divide by avogadro's number since it is asking for energy per sodium atom. Finally for C, I believe you would use E=hv to find the frequency.

Jiapeng Han 1C
Posts: 85
Joined: Wed Sep 30, 2020 9:50 pm

Re: Photoelectric Effect

Postby Jiapeng Han 1C » Thu Oct 15, 2020 9:12 pm

For part A: kinetic energy is just Ek=1/2 m v^2
For part B: note the unit here--Kj/mol, so you need to divide the work function by the Avogadro's constant, and the unit should be Jk
For part C: firstly you add work function and kinetic energy together. Then you divide the sum by the Planck constant.


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