Photoelectric Effect Module Question

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Elizabeth Kaplan 3I
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Photoelectric Effect Module Question

Postby Elizabeth Kaplan 3I » Tue Oct 20, 2020 10:49 pm

33. Molybdenum metal must absorb radiation with a minimum frequency of 1.09 x 1015 s-1 before it can emit an electron from its surface. Answer the following two questions.
A. What is the minimum energy needed to produce this effect?

34. B. If molybdenum is irradiated with 194 nm light, what is the maximum possible kinetic energy of the emitted electrons?

I understand how to do 33A, however I cannot seem to get the correct answer for 34B.

Stuti Pradhan 2J
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Re: Photoelectric Effect Module Question

Postby Stuti Pradhan 2J » Tue Oct 20, 2020 10:56 pm

Since the kinetic energy is the difference between the energy of the photon and the energy required to emit a photon, you should figure out the energy that corresponds to a wavelength of 194 nm to begin 34B. Subtract the energy you found in 33A (the work function for Molybdenum) from the new energy that you just found, to find the maximum kinetic energy.

Hope this helps!

Stacey Phan 2I
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Re: Photoelectric Effect Module Question

Postby Stacey Phan 2I » Fri Oct 23, 2020 4:14 am

Hi so I'm going to explain part A and part B in case anyone is wondering how to do part A as well.

Part A)
You are given a frequency of 1.09E15 Hz so you want to use an equation that involves energy and frequency. Use equation E=hv.
E=(6.626E-34 Js) * (1.09E15 Hz) = 7.22234E-19 J. This is the threshold energy or work function (minimum energy needed to emit an electron). Based on the frequency given, there are 3 sig figs so your answer should be 7.22E-19 J.

Part B)
The incoming light has a wavelength of 194 nm so convert that to meters (1.94E-7 m). You are given a wavelength so you want to use an equation that involves energy and wavelength. You can manipulate the c=lambda*frequency to substitute it for the frequency in E=hv. E=h*(c/lambda).
E=(6.626E-34 Js) * (3E8 m/s / 1.94E-7m) = 1.02463918E-18 J. This is the energy of a photon.

To solve for the kinetic energy use the equation E(photon) = work function + E(kinetic).
Rearrange the equation to solve for E(kinetic). E(kinetic) = E(photon) - work function
E(kinetic) = (1.02463918E-18 J - 7.22234E-19 J) = 3.02405175E-19 J (PS I wouldn't use rounded numbers until the end of solution so you can get exact answer)
Based on the frequency given, there are 3 sig figs so your answer should be 3.02E-19 J.

Hope this helps! :)

Darren1j
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Re: Photoelectric Effect Module Question

Postby Darren1j » Sun Oct 25, 2020 5:59 pm

Hey, I have question from this same module. Question 30. "Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1. .... What is the frequency of the incident light on the sodium metal surface?"

Are you supposed to convert KJ per mole to Joule per atom for this question?

Andy Hernandez
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Joined: Wed Sep 30, 2020 9:50 pm

Re: Photoelectric Effect Module Question

Postby Andy Hernandez » Sun Oct 25, 2020 9:39 pm

Darren1e wrote:Hey, I have question from this same module. Question 30. "Light hits a sodium metal surface and the velocity of the ejected electron is 6.61 x 105 m.s-1. The work function for sodium is 150.6 kJ.mol-1. .... What is the frequency of the incident light on the sodium metal surface?"

Are you supposed to convert KJ per mole to Joule per atom for this question?

yes, so itll be 150600 (1 mol/ 6.022x10^23)= 2.501x10^-19
then, to find the frequency, u rearrange, E=hv to v=E/h.
u plug in, 1.99x10^-19 (kinetic energy) + 2.501x10^-19 (work function aka threshold energy) for E. Divide by H.
Answer is, 6.78x10^14


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