## hw problem 1.33 c

Lindsay H 2B
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Joined: Fri Sep 29, 2017 7:07 am
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### hw problem 1.33 c

The velocity of an electron that is emitted from a metallic surface by a photon is 3.6x10^3 km.s-1.
(a) What is the wavelength of the ejected electron?
(b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50x10^16 Hz. How much energy
is required to remove the electron from the metal surface?
(c) What is the wavelength of the radiation that caused photoejection of the electron?
(d) What kind of electromagnetic radiation was used?

I don't understand why for part C, I can't just use wavelength=c/frequency and the frequency given in part B to find the wavelength of the radiation, since both B and C are only asking about the energy required to emit an electron, not the kinetic energy of the electron afterwards. The solutions manual shows that we have to find the kinetic energy as well, that's why im confused.

Ryan Sydney Beyer 2B
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### Re: hw problem 1.33 c

For part C, the question is asking for the wavelength of the electromagnetic radiation that hit the metallic surface. We also know that the energy of a singular incoming photon, E, is equal to the work function added to the kinetic energy of the electron. This is because the energy that was put into the metallic surface is used towards the work function first and then any excess is in the kinetic energy. This equation can be seen as :

$h\nu = \phi + 1/2mv^2$

So we know that this electron has kinetic energy because its velocity is given to us in the problem. Also we can calculate the energy of the work function since the problem also provides us the frequency of the radiation.

Rachel Wang
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

### Re: hw problem 1.33 c

You find in part a) the kinetic energy of the ejected e-: 5.909x10^-18 J.
You find in part b) the work function, aka threshold energy because it's the minimum required energy to eject the electron, not the incident light.

To find the energy of the incident light,
E - 1.658x10^-17 = 5.909x10^=18
E = 2.249x10^-17
To find the λ
E = hc/λ
λ=8.84x10^-9m

Hope this helped! C: