6th Edition 1.33 (c)

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305144105
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Joined: Thu May 10, 2018 3:00 am

6th Edition 1.33 (c)

Postby 305144105 » Mon Oct 22, 2018 6:36 pm

1.33 The velocity of an electron that is emitted from ametallic surface by a photon is 3.6*10^3 km s^-1.
(c) What is the wavelength of the radiation that caused photoejection of the electron?

Can someone please explain what formula to use and why?

Mahir_Hasan2C
Posts: 60
Joined: Fri Sep 28, 2018 12:25 am

Re: 6th Edition 1.33 (c)

Postby Mahir_Hasan2C » Mon Oct 22, 2018 7:26 pm

In the previous question, as in b, we found the threshold energy. What we currently have is the threshold energy, the mass of the electron (which is implied when it states electron), the velocity of the electron (which we converted to m/s). With the mass and velocity of the electron, we can find the kinetic energy of the electron. Through the equation hv-threshold energy=kinetic energy, we can find hv (incoming energy) through adding threshold energy and kinetic energy. We can then divide h to both sides and this would be your frequency. We can use this frequency to solve for wavelength in the equation c=lambda*v. Then we solve for lambda convert to nm (with significant figures). This would be your answer.


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