[jtlbb2]

Okay, I have a problem that I am stumped on:

"[Delta]E = -2.18 * 10(-18) J/atom (1/n^2final - 1/n^2initial)

The Lyman series corresponds to transitions from excited electronic states to a final state of n = 1.  If the wavelength of a transition in the Lyman series is 102.5nm, what is the initial state of the atom?"

I'm not even sure if I am on the right track, but this is what I do to start the problem:

C = (wavelength)(frequency)

2.998 * 10^8 m * s^(-1) = 102.5nm * V

102.5nm * 1m/1.0*10^9nm = 1.0 * 10^(-7).

V = 2.998 * 10^8 m * s^(-1)/(1.0 * 10^-7 m)

V = 2.998 * 10^15 * s^(-1)

E = hv

E = (6.626 * 10^(-34) J*s)(2.998 * 10^15 * s^(-1))

E = 1.99 * 10^(-18) J

And this is as far as I can get.  I don't even know if this is the right idea.  If this is the right track, then where do I go from here?  If not, then please set me straight

[plu]

102.5nm * 1m/1.0*10^9nm = 1.0 * 10^(-7).

Check your calculations here.  Then, knowing that the "final state of n = 1" (substitute 1 for n_final), you can find n_initial simply with algebra

[jtlbb2]

I still can't solve it algebraically.  Am I on the right track with finding the energy?  I'm not sure what to do with the value of energy once I have calculated it.

[plu]

Once you have your value for energy, you will only have one unknown variable left in your equation (n_final).

[jtlbb2]

The n²_final is known, it's the initial that isn't known.  Besides, this isn't helping me to answer the question.  As I've said, I don't know what to do with the value for energy that I have calculated.  Where do I plug this into the problem?  If this is done by simple algebra, then I am totally missing how to do that.

[Donaldson Tan]

1/lamda = deltaE/h.c

[plu]

"[Delta]E = -2.18 * 10(-18) J/atom (1/n^2final - 1/n^2initial)

Your equation is right here!  You have the dE value (the energy value you calculated) and you have the n_final value.  Then, you only have one variable left.  I'm sorry but I don't quite understand where the problem lies  

[jtlbb2]

Well, I have shown how I approached the problem and have asked if I am even on the right track.  I never received a straight answer on that.  Since you're now saying that all I need to do to work the problem is "[Delta]E = -2.18 * 10^(-18) J/atom (1/n^2final) - (1/n^2initial), I'm going to assume that I was on the wrong track with how I approached the problem in my very first post.

But let's see what happens if I try to solve for n^2initial algebraically.

(-2.18 * 10^(-18)) J *atom^(-1))/1 - ((-2.18 * 10^(-18))/n^2) = 0

(-2.18 * 10^(-18)) J * atom^(-1)n^2)/n^2 - (-2.18 * 10^(-18) J * atom^(-1))/n^2 = 0

(-2.18 * 10^(-18))J * atom^(-1))(n^2) + 2.18 * 10^(-18) J * atom^(-1) = 0

(-2.18 * 10^(-18))J * atom^(-1))(n^2) = -2.18 * 10^(-18) J * atom^(-1)

n^2 = 1

n = + or - 1.

Okay, assuming that I did not make a mistake somewhere, I get n = 1.  But this isn't the right answer.  I already know the answer to this problem, I just can't get it.  

I've also asked some of my other classmates to work this problem algebraically, and they couldn't get the right answer either.  So I'm either making a mistake somewhere, or I am approaching this problem wrong.

geodome,
I'll try to work this problem with the R value that you provided, and see if it gets me the correct answer.  Out of curiousity though, why are you throwing out the "-2.18 * 10^(-18)" and replacing it with R?

[Donaldson Tan]

jtlbb2:

i have no idea how you all do your calculations, because your original question gives me the value of 3.

dE = -2.18E-18)(1-n^-2)

since the wavelength emmitted is 102.5nm, then
dE = -hc/lamda = -(6.626E-34)(3E8)/(102.5E-9) = -1.9393E-18 J/atom

=> 1.9393E-18 =( 2.18E-18)(1-n^-2)

(1-n^-2) = 1.9393E-18/2.18E-18 = 0.8896

n^-2 = 1- 0.8896 = 0.110396
n² = 9.0583 => n = 3

[jtlbb2]

Unfortunately, 3 is not the correct answer. But thanks for giving it a shot.  

[Donaldson Tan]

If I am wrong, what is the correct answer?

If the answer is not 3, then are you sure this involves the lyman seriess?

[jtlbb2]

That problem is quoted directly from the worksheet that I received.  I didn't add or take anything away from it.  I just typed it as it was given.  And the answer is 4.

Here is the question, again, in its entirety incase there was any confusion:

"[Delta]E = -2.18 * 10(-18) J/atom (1/n^2final - 1/n^2initial)

The Lyman series corresponds to transitions from excited electronic states to a final state of n = 1.  If the wavelength of a transition in the Lyman series is 102.5nm, what is the initial state of the atom?"

[Donaldson Tan]

your model answer is incorrect.