The velocity of an electron that is emitted from a metallic surface by a photon is 3.6*10^3km/s
(b) No electrons are emitted from the surface of the metal until the frequency of the radiation reaches 2.50*10^16Hz. How much energy is required to remove the electron from the metal surface?
I was able to solve part a of this problem, but how do I go about part b?
Textbook 1B #15
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Re: Textbook 1B #15
To remove an electron from the surface the metal, the energy of the photon must be greater than or equal to the work function. In this case, you would use E=hf to find the energy of the photon that is able to remove the electron.
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Re: Textbook 1B #15
Hi!
For part (b), you can use the equation E = (h)(frequency) in order to find the energy required to remove the electron from the metal surface. This problem is basically asking you to find the work function. Since it states that no electrons are emitted until the frequency reaches 2.50 x 10^16 Hz, this can be used to find the work function.
E = (h)(frequency) = (6.63 x 10^-34 Js)(2.50 x 10^16 s-1) = 1.66 x10^-17 J
Hope this helped! :D
For part (b), you can use the equation E = (h)(frequency) in order to find the energy required to remove the electron from the metal surface. This problem is basically asking you to find the work function. Since it states that no electrons are emitted until the frequency reaches 2.50 x 10^16 Hz, this can be used to find the work function.
E = (h)(frequency) = (6.63 x 10^-34 Js)(2.50 x 10^16 s-1) = 1.66 x10^-17 J
Hope this helped! :D
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Re: Textbook 1B #15
For part B I used E = hv so then you plug it all in with 6.636x10^-34(2.5x10^16) to get the answer for this part
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Re: Textbook 1B #15
For part b you would need to set the kinetic energy equal to 0, and then you would need to equate the energy to the work function of the atom. Hope this helps!
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