Question 6, Quiz 1, Section 3F

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Coco Hailey 2E
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Joined: Fri Sep 25, 2015 3:00 am

Question 6, Quiz 1, Section 3F

Postby Coco Hailey 2E » Sun Nov 01, 2015 5:11 pm

The Photoelectric effect is the observation that many metal emit electrons when light shines upon them. Given the threshold energy of copper (7.53x10^-19J), calculate the velocity in (m/s) of an electron that is emitted when electromagnetic radiation with a wavelength of 161nm is shone on the surface.

Annie Qing 2F
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

Re: Question 6, Quiz 1, Section 3F

Postby Annie Qing 2F » Sun Nov 01, 2015 8:07 pm

I would like to preface this with a "sorry the equation editor isn't working for me."
Anyways,


From E=h(frequency) and c=(wavelength)(frequency), we can get E=h(c/wavelength).

Given the wavelength of the emitted light is 161x10^-9 m, c=3.00x10^8 m/s, and h=6.62608x10^-34 J•s,
E=h(c/wavelength)
E=(6.62608x10^-34 J•s)(3.00x10^8 m/s)/(161x10^-9 m)
E=1.23x10^-18 J
Therefore, the total energy of the emitted light is 1.23x10^18 J.

Let E = total energy of the emitted light (same E as before)
E = energy used to eject the electron, threshold energy
E = kinetic energy of the electron
m = mass of an electron
v = velocity of the electron after ejection

Since the energy of the emitted light goes into ejecting the electron and then the remaining energy is used in the kinetic energy of the photon,
E = E + E
1.23x10^-18 J = 7.53x10^-19 J + (1/2)mv^2
4.77x10^-19 J = (1/2)(9.1095x10^-31 kg)(v^2)
1.05x10^12 m^2/s^2 = v^2
v = 1.02 x 10^6 m/s

Therefore, when the electromagnetic radiation with wavelength 161 nm is shone, the electron emitted has velocity 1.02x10^6 m/s.


Let me know if you have any questions about what I did!


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