module question 28 a

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

module question 28 a

Postby melissa_dis4K » Sun Oct 21, 2018 11:36 pm

For this question, "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 kinetic energy of the ejected electron?" I know that E(photon)-E(energy remove e-)=Kinetic energy of the electron.
h*frequency(v)-work function=1/2MeVe^2. I plugged in h=6.626 x 10^-34 Js and the given work function 150.6 kJ.mol-1. I also know that the velocity is 6.61 x 10^5 m.s-1. I am confused as where we get the mass of the electron from will it be given? And also once I do have that do I just ignore the rest of the equation and only use Ek=1/2Me*Ve^2? If so, when would I use the given work function then?

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

Re: module question 28 a

Postby Heesu_Kim_1F » Mon Oct 22, 2018 1:48 am

Problem 28A is only asking for you to solve for the kinetic energy. Thus, the only equation to use for this problem is (1/2)mv^2. Velocity is given in the problem: 6.61 x 10^5 m/s, and the mass of electron is a number/constant that will be given on the equation sheet: 9.11 x 10^-31 kg. Plug those numbers, and I got 1.99 x 10^-19 J. The photoelectric equation, E(photon)-E(energy remove e-)=Kinetic energy of the electron OR h*frequency(v)-work function=1/2MeVe^2, comes into play for problem 30C. Hope this helps!


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