Solve Electron Velocity to Find Initial Energy State

[usfz28]

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))

Does that sound about right? Sorry about all the parenthesis

 

[usfz28]

anyone??

 

[dextercioby]

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

Daniel.

 

[usfz28]

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=(Ke^2/2a_{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)

 

[usfz28]

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 couldn't figure it out

 

[usfz28]

Hope you understand

 

[James R]

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}