Photoelectric Effect

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Joey_Okumura_1E
Posts: 122
Joined: Wed Sep 30, 2020 9:58 pm

Photoelectric Effect

Postby Joey_Okumura_1E » Sat Dec 12, 2020 9:55 pm

Can someone please explain how to do this problem?: A newly designed laser pointer with a certain frequency is pointed at a sodium metal surface. An electron is ejected from the metal surface with wavelength 1.10 nm. What is the frequency of the light from the laser pointer? The work function of sodium is 150.6 kJ∙mol-1

Mohamed Mido
Posts: 140
Joined: Wed Sep 30, 2020 9:33 pm

Re: Photoelectric Effect

Postby Mohamed Mido » Sat Dec 12, 2020 10:05 pm

First you'll have to figure out the kinetic energy of the electron from the wavelength they gave you. You're going to plug it into λ=h/mv to find out the velocity. Then you'll plug in the velocity into E=1/2mv^2. Then you will add that energy to the work function they provided. Then find out the frequency of the light by plugging the energy value in E=hv, rearranging the equation to v=E/h.

abby hyman
Posts: 131
Joined: Thu Jul 25, 2019 12:16 am

Re: Photoelectric Effect

Postby abby hyman » Sat Dec 12, 2020 10:16 pm

To begin, you would need to use the wavelength given (converted to meters) to find the velocity of the ejected electron.
The equation for this would be deBroglie's so wavelength=planks constant/(mass)(velocity)
Once you have the velocity of the electron, you can use 1/2mv^2 to find the Kinetic energy of the ejected electron
Next, you must convert the work function from kJ/mol to Joules
To do this you would have to multiply 150 by 1000 and divide by avagadros number
Now you can add the Kinetic energy and the work function to find the energy of the photon
Once you have the energy of the photon you can set it equal to hv and divide by planks constant to find the frequency.

When I did it I got 6.765x10^14

Madeline Louie 1I
Posts: 56
Joined: Wed Sep 30, 2020 9:48 pm

Re: Photoelectric Effect

Postby Madeline Louie 1I » Sat Dec 12, 2020 10:18 pm

Mohamed explained it really clearly, and for questions like this, you have to remember that the E=hv and c=λv only apply to electromagnetic radiation. Therefore, you can't use these equations with an electron when trying to find its wavelength or energy. That is why you need to use de Broglie's equation to get the velocity of the electron before its kinetic energy.


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