Post Module Assessment - Photoelectric Effect

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Ghadir Seder 1G
Posts: 134
Joined: Sat Sep 14, 2019 12:17 am

Post Module Assessment - Photoelectric Effect

Postby Ghadir Seder 1G » Wed Oct 02, 2019 10:59 pm

So I was doing the post module assessment for the the photoelectric effect and I encountered this question...

Calculate the energy per motion of ultraviolet radiation of frequency 3.00x10^15Hz.

I used the formula : E=h(wavelength) and got the right amount (1.99) but the "x10" was raised to the power of 50, rather than -18, which is the correct answer.

Any ideas as to what I may have done wrong?

Thank you !

Victoria Zheng--2F
Posts: 103
Joined: Fri Aug 09, 2019 12:17 am

Re: Post Module Assessment - Photoelectric Effect

Postby Victoria Zheng--2F » Thu Oct 03, 2019 8:57 am

Hello. I think you might have accidentally input the wrong power of 10 in the calculator. To do this question, you would use the equation E=hv(energy= to Planck's constant*frequency), so when you plug in the numbers into the equation, you would have E=(6.626*10^-34)*(3.00*10^15), which will give you the number1.9878*10^-18, and since the question gives 3 significant figures, the answer would be 1.99*10^-18. I hope this helps.

Ghadir Seder 1G
Posts: 134
Joined: Sat Sep 14, 2019 12:17 am

Re: Post Module Assessment - Photoelectric Effect

Postby Ghadir Seder 1G » Thu Oct 03, 2019 9:50 am

This helped so much!


Thanks!

Claire Stoecklein 1E
Posts: 49
Joined: Sat Jul 20, 2019 12:15 am

Re: Post Module Assessment - Photoelectric Effect

Postby Claire Stoecklein 1E » Mon Oct 07, 2019 3:32 pm

When I use scientific notation in my calculator, I make sure to use parentheses around each answer so the numbers don't get confused. For example just put (1.99 x 10^-18) in the calculator.


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