Test 2 #7  [ENDORSED]

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Adela Henry 1I
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Test 2 #7

Postby Adela Henry 1I » Fri Apr 27, 2018 4:20 pm

Could someone explain how to do #7 from test 2
Thanks

princessturner1G
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

Re: Test 2 #7  [ENDORSED]

Postby princessturner1G » Fri Apr 27, 2018 4:50 pm

Adela Henry 1I wrote:Could someone explain how to do #7 from test 2
Thanks

You use the equation Energy of a photon-work function=Kinetic Energy. Since it is asking for the maximum wavelength kinetic energy equals 0. The energy of the photon is then equal to the work function. The energy of the photon= 9.02 x 10^-19 J. Since the energy of the photon equals Planck's constant x frequency, you can then solve for the wavelength.

AshleyLamba1H
Posts: 30
Joined: Fri Apr 06, 2018 11:03 am

Re: Test 2 #7

Postby AshleyLamba1H » Sun Apr 29, 2018 12:37 am

Could you post a photo of the question please? That would be helpful for people to answer your question.

Eli Esagoff_1I
Posts: 38
Joined: Fri Apr 06, 2018 11:02 am

Re: Test 2 #7

Postby Eli Esagoff_1I » Sun Apr 29, 2018 1:42 pm

For this question you had to use the equation: (Energy photon - threshold energy = kinetic energy). The kinetic energy is 0 so you just do (hc)/wavelength = the given threshold energy. After plugging in variable, you can sold for wavelength which equals 2.20x10^-7 m. Hope this helped.


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