Exam 2 Question 4A

[Hannah Faris 1C]

The question was : at the threshold voltage, light is emitted with energy equal to 3.61*10^-22kJ, what is the frequency of the photon?

I used both of the light equations to convert the energy into the frequency, but I'm not entirely sure what I did wrong. I would be really thankful if I could see someone else's work so I can see where I went wrong!

[Schem_student]

You must use E= freq x h but make sure to convert KJ to joules because planc's constant is in J x s.

[Sophia Ding 1B]

Which two light equations did you use? Because based on the wording of the question, I think only E = hv would need to be used as you are already given the energy of the photon. So you would want to convert the 3.61 x 10^-22 kJ into J (*1000), and then set that equal to hv, where h is equal to Planck's constant. Thus, solving for v would be E/h, or (3.61 x 10^-19 J)/(6.626 * 10^-34 Js).

[anishathomas]

You have to use e=hv as thst is the energy of the photon. Since you have the energy already you just have to solve for frequency as h is already given.

[Ian Marquez 2K]

As stated above, e=hv would be sufficient for this problem. The only thing you have to watch out for is converting kJ (10^3) to J. Other than that, this is the energy of a single photon so you would just multiply this number with Planck's constant and apply proper significant figures.

[shaunajava2e]

e=hv, like stated above should have been used, but it was also necessary to convert the units from kJ to J which could result in error as well.

[Archana Biju 1G]

You have to convert from kilojoules to joules, since the h constants unit is j.s, so that they cancel out when you use the equation e = hv, v = e/h, to get the frequency in the unit s^-1.