Photoelectric Effect Module Question 27

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melissa_dis4K
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Joined: Fri Sep 28, 2018 12:28 am

Photoelectric Effect Module Question 27

Postby melissa_dis4K » Sun Oct 21, 2018 11:17 pm

For question 27 it asks "Calculate the energy per photon of ultraviolet radiation of frequency 3.00 x 1015 Hz." What I did was, using E=h*frequency(v) and c=lambda*frequency(v) I solved for E by rearranging the equations to give E=hc/lambda =(6.626*10^-34 Js)(3.00 x 10^8 ms^-1)/3.00 x 10^15 Hz and then I multiplied this by Avogadro's number 6.022 x 10^23 to get photons. Does anyone know if I am doing this wrong because I get the wrong answer: -1.83 x 10^-23.

Mukil_Pari_2I
Posts: 87
Joined: Fri Sep 28, 2018 12:29 am

Re: Photoelectric Effect Module Question 27

Postby Mukil_Pari_2I » Sun Oct 21, 2018 11:32 pm

all you need to do is multiply the frequency by plancks constant. The correct answer if you just do hv is 1.99*10^-18 J.

Elias Omar 1G
Posts: 52
Joined: Fri Sep 28, 2018 12:17 am

Re: Photoelectric Effect Module Question 27

Postby Elias Omar 1G » Sun Oct 21, 2018 11:37 pm

In order to find the energy per photon, you just need to know the frequency of the radiation. You only need to use E=h*v. This formula gives you the energy per photon. The answer you should get is around 19.88*10^-19 (E= h(*6.626*10^-34) * v(3.00*10^15). Or E=1.99*10^-18.


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