Module Questions 33 and 34  [ENDORSED]

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Or Fisher 1I
Posts: 35
Joined: Fri Feb 02, 2018 3:00 am

Module Questions 33 and 34

Postby Or Fisher 1I » Mon Apr 23, 2018 4:37 pm

Hello, I managed to figure out 33 was 7.22x10-19J, but I do not understand how to then use this to solve 34. Can someone please help me? Thanks.
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?

A. 4.51 J

B. 7.22 x 10-12 J

C. 7.22 x 10-19 J

D. 1.15 x 10-19 J

E. None of the above
1/1 point

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

A. 6.41 J

B. 3.05 x 10-19 J

C. 1.03 x 10-11 J

D. -5.58 x 10-19 J

E. None of the above

RubyLake1F
Posts: 41
Joined: Fri Apr 06, 2018 11:03 am

Re: Module Questions 33 and 34  [ENDORSED]

Postby RubyLake1F » Mon Apr 23, 2018 6:51 pm

In question 33 you found the minimum possible energy of light that will eject an electron (ejecting the electron, but leaving it with a kinetic energy of zero), and in question 34 you are given the wavelength of the actual photons of light hitting the source (implying that there is extra leftover energy that will result in kinetic energy in the ejected electrons). You need to convert the wavelength of the incident light (194 x 10^-9 meters) into energy using the equations wavelength x frequency= speed of light, and frequency x Planck's constant=Energy.
Next, you can find the "leftover" energy (which will equal the energy of the ejected electron) by subtracting the work function (required energy to eject electron) from the energy of the photon that the material was hit with.
Hopefully this makes sense! If it doesn't I can type up the numbers as well :)

jessicasam
Posts: 36
Joined: Wed Nov 15, 2017 3:02 am
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Re: Module Questions 33 and 34

Postby jessicasam » Wed Apr 25, 2018 9:40 pm

Can you explain the part about subtracting the ""leftover" energy (which will equal the energy of the ejected electron) by subtracting the work function (required energy to eject electron) from the energy of the photon that the material was hit with". Numbers might help. Thanks!

RubyLake1F
Posts: 41
Joined: Fri Apr 06, 2018 11:03 am

Re: Module Questions 33 and 34

Postby RubyLake1F » Mon Apr 30, 2018 12:30 pm

The equation for the photoelectric effect is this:

(Energy of incoming photon) - (energy required to eject electron from surface) = (kinetic energy of ejected electron)

This makes sense because there is conservation of energy in all reactions, so whatever energy is leftover from ejecting the electron will be converted into kinetic energy of the ejected electron.

In this case, the work function (energy required to eject electron) is 7.22x10-19J. You are given the wavelength of the photon that is hitting the surface, and are asked to find the maximum possible "leftover" energy that will be turned into kinetic energy of the ejected electron. Converting the wavelength of the incoming photon into Joules using the equations I mentioned in my previous comment, you can find that its energy is 1.0245 x 10-18 J.
Now set up the photoelectric equation:

1.0245 x 10-18 J (energy of incoming photon) - 7.22x10-19 J (energy required to eject electron)= excess energy (kinetic energy of electron).

When you subtract the answer will come out to 3.05 x10-19J, which is the maximum possible kinetic energy of an ejected electron given the wavelength of the photon it was ejected by.

The answer is B :)


Sorry for the late response, hope this helps!


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