Minimum potential difference, Balmer series

[robbie8292]

Homework Statement

-A beam of electrons bombards a sample of hydrogen in its ground state.
a) Through what minimum potental difference must the electrons be accelerated if only one line of the Balmer's series can be observed?

Homework Equations

V= h^2/2meλ^2 (V=potential difference)
1/λ=R((1/n(i)^2)-(1/n(f)^2))

The Attempt at a Solution

...1/(1.09*10^7((1/2^2)-(1/3^2)))=λ, λ=6.60*10^-7
V= h^2/2meλ^2
...((6.626*10^-34)^2/((2*(9.11*10^-31)(1.6*10^-19)(6.60*10^-7)^2))= 3*10^-6 volts
[/B]
This is a question I got wrong on a previous homework and I am still trying to figure out how to correctly figure this out. I just am unsure if I am even doing this correctly and I have tried this in quite a few ways.

 

[TSny]

Hello. Can you explain why you chose n(i) = 2 and n(f) = 3 ? Note that the atoms are initially in the ground state. The electrons must excite the atoms from their ground state to the appropriate energy level for which the first Balmer line can subsequently be produced.

 

[robbie8292]

I chose 2 because that is where the Balmer lines start so that is what I assumed I must use. If that is wrong may you elaborate on where I should begin at?

 

[TSny]

The atom cannot emit light unless it is first excited into an excited state. Where does the energy come from to excite the atom? Also, to what minimum energy level must the atom be excited in order to be able to emit one of the Balmer lines?

 

[robbie8292]

The first excited state is -13.6Ev/4= -3.4eV is where the Balmer lines start due to the n=2 is the point after the ground state. Also you excite the atom through collisions to bump one from one state to another. Would I use E = q*V and just subtract the first state from the 2nd? Such as -13.6Ev/1.6*10^-19C - -3.4 eV/1.6*10^-19C= -6.38 *10^19 Volts would be the potential difference?

 

[gneill]

You need to look into what the Balmer series represents (I think there's a Wikipedia article that explains the various series fairly well, if memory serves). In short, the series is a result of electrons jumping down to the 2nd state. In order to jump down to the second state they must first be kicked up to a higher state than that...