# Photon Energy Levels

My instructor was telling what would be on the upcoming test and he said something about:Given the velocity of an electron, the work function of a certain metal, and final energy level. We should be able to find the intial energy state. Sound pretty easy.... to easy but here is what i was thinking...
Given the velocity of the electron, I can found out the kinetic energy of the electron 1/2MV^2=K.E.
With the K.E. I can then use f=((work funct)+(K.E.))/(H) to find the Freq. () With that I can then find λ=(C)/(F). to find λ the wavelength.
Then I finally can find the initial energy state by using:
N(initial)=Sq Root(((k(e)^2)/2(Aof zero)hc)-1/λ+n(final)^2))

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anyone?? dextercioby
Homework Helper
Yes,me,it looks okay.What does the last fomula represent...?Could u write it using LaTex...?

Daniel.

Ok ill try.Its the Blamer formula. This equation he gave us. It's not in the book it has to do with the bohr model. This is the way he gave it to us
1/λ=(K$$e^2$$/2$$a_{0}$$hc)(1/$$n_{f}$$$$^2$$ -1/$$n_{i}$$$$^2$$)

Where K=coloumb's constant
e=charge of electron
$$a_{0}$$= lowest orbit radia (what you get when $$r_{n}$$=1 bohr atom radi of orbit)
H=Planck's constant
c=speed of light
All those are known Rydberg constant
$$n_{f}$$=final energy level
$$n_{i}$$=initial enrgy level (this is what we are suppose to find)

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N(initial)=Sq Root(((k(e)^2)/2(Aof zero)hc)-1/λ+n(final)^2))

i ended up with this

$$n_{i}=all sqroot(ke^2/2a_{0}$$hc - $$\frac{\1}{\lambda} + n_{f}^2$$)

There is supposed to be a 1 over the lambda but couldnt figure it out

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Hope you understand

James R
Homework Helper
Gold Member
usfz28:

Click on any equation to see the LaTeX code you need.

In the above case, it is:

$$N_i=\sqrt{\frac{ke^2}{2 a_0 hc} - \frac{1}{\lambda} + n_f^2}$$