# Electron Energy Level Fall at a Given Wave Length

1. Nov 17, 2015

### ptownbro

1. The problem statement, all variables and given/known data

An electron in the H atom falls from n=(?) to n=3. The wave length of the emitted photon is 1100 nm. From what level did the photon fall?

2. Relevant equations

Used the following to get the Energy.

w = c / v
E = h * v
E = h * (c/w)

w = lambda (couldn’t figure out how to use the real lambda symbol)
c = speed of light at 3*10^8 m/s (where “m” is meters and “s” is seconds)
v = frequency in Hz
E = Energy in Joules
h = planks constant at 6.62607*10^-34*J*s (where “J” is joules and “s” is seconds)

Then used the following along with the Energy calculated from above

Change in Energy = -2.18*10^-18 * (1/n2^2 – 1/n1^2)

3. The attempt at a solution

a) Converted 1100nm to m.

=11*10^-7 m

b) Calculated frequency using w = c / v

v = (3*10^8 m/s) / (11*10^-7 m)
v = (3*10^8 s^-1 * 10^7) / 11
v = (3*10^15 s^-1) / 11
v = 2.73*10^14 s^-1

c) Calculated Energy using E = h * v (with v calculated in part b)

E = (6.62607*10^-34*J*s) * (2.73*10^14 s^-1)
E = (6.62607*10^-20*J) * (2.73)
E = 18.07*10^-20 J

d) Finally, calculated the level using the Energy

18.07*10^-20 J = (-2.18*10^-18) * (1/3^2 – 1/x^2)
(1/9 – 1/x^2) = (18.07*10^-20 J) / (-2.18*10^-18)
(– 1/x^2) = (18.07*10^-20 J) / (-2.18*10^-18) * 9
x^2 = - (-2.18*10^-18) / (18.07*10^-20 J * 9)
x = [(2.18*10^-18) / (-18.07*10^-20 J * 9)] ^ 1/2
.....

Stopped because it didn’t seem right =)

2. Nov 17, 2015

### Staff: Mentor

In the first line above, be careful with the sign. What energy should appear on the left-hand side: the energy of the photon or of the atom?

You have an error going from the second line to the third.

3. Nov 17, 2015

### ptownbro

The energy of the photon.

While I fix my Algebra, can you tell me if we're on the right track and if our thinking is correct?

4. Nov 17, 2015

### Staff: Mentor

Is that compatible with your equation?
Apart from the sign problem I'm alluding to, yes.

5. Nov 17, 2015

### ptownbro

I got the "Change in Energy" equation from the following youtube video:

I also understood and assumed that "Change in Energy" is the "amount of energy" released when the electrons change levels. So to me it made sense to make those equal to each other. So, re-written, it should read as:

-2.18*10^-18 * (1/n2^2 – 1/n1^2)

If my energy is

E = 18.07*10^-20 J

Then in my part d above I should be able to do:

18.07*10^-20 J = (-2.18*10^-18) * (1/3^2 – 1/x^2)

Where I try to solve for "x".

(Still working on correcting my math).

6. Nov 17, 2015

### ptownbro

So I've reworked my math and here's where I'm at now. Again, the question is:

An electron in the H atom falls from n=(?) to n=3. The wave length of the emitted photon is 1100 nm. From what level did the photon fall?

From my original work above, I was able to get from 1100 nm to figure out the amount of energy that was released during the electron fall from n=(?) to n=3 as being equal to:

E = 18.07*10^-20 J

Then, using the following equation (given from the video referenced above (and partially confirmed from my daughter's notes) (By the way, I erroneously left out the "J" in the equations above. sorry!) :

E = -2.18*10^-18 J * (1/n2^2 – 1/n1^2)

Therefore, I should be able to solve for n1 (which I've replaced with "x") as follows:

18.07*10^-20 J = (-2.18*10^-18 J) * (1/(3^2) - 1/(x^2))
(18.07*10^-20 J) / (-2.18*10^-18 J) = (1/(9) - 1/(x^2))
(0.1807*10^-18 J) / (-2.18*10^-18 J) = (0.1111 - 1/x^2) <--- here i just simply moved the decimal in the first pat to get both exponents to -18 so they can cancel
(0.1807) / (-2.18) = (0.1111 - 1/x^2)
-0.0829 = 0.1111 - 1/x^2
-0.0829 - 0.1111 = - 1/x^2
x^2 = -1 / -0.1940
x = 5.1546^(1/2)
x = 2.27

So.... by my calculations the electron from the H atom fell from level n=2.27 to n=3.

Is that correct?

Seems weird that a level is not a whole number? Hmm...

7. Nov 17, 2015

### Staff: Mentor

I've looked at the video, and I think that it makes things very confusion for her to use the notation $\Delta E_\mathrm{photon}$, as that is the change in energy of the atom, not the photon. She seems to be using the label photon because it is the "energy per photon" (which is also a strange thing to say).

So, lets try to do this from scratch. The energy of an orbit in the Bohr atom is $E_n = -R / n^2$, where $R$ is the Rydberg constant expressed in units of energy. In SI units, you have $R = -2.18 \times 10^{-18}\ \mathrm{J}$. Note that the Rydberg constant is often expressed in units of inverse length, because Rydberg originally used his formula to calculate the inverse wavelength $1 / \lambda$.

For a transition from $n_2$ to $n_1$, we get, for the atom,
$$\Delta E = E_\mathrm{final} - E_\mathrm{initial} = \frac{-R}{n_2^2} - \frac{-R}{n_1^2} = -R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right)$$
Going from a higher energy state to a lower energy state, you should have $\Delta E < 0$.

Try again with the correct sign, and you should get a value of $n$ that is an integer (within rounding errors), and which is greater than 3, as the transition is from a higher excited state down to $n=3$.

Last edited: Nov 17, 2015
8. Nov 17, 2015

### ptownbro

I'm sorry, but I don't see what you mean here? I don't see a difference in a sign as you indicated.

9. Nov 17, 2015

### Staff: Mentor

Sorry, I had made a mistake in post #7. I have now edited that post.

I get (-0.1807) / (-2.18) = (0.1111 - 1/x^2). The sign on the left-hand side is not the same, because since it is $\Delta E$ for the atom, it is negative.

10. Nov 17, 2015

### ptownbro

Hm. Why would it be that way? It's falling from n=? to n=3.

I read that is n=? is the starting point (or n1) and n=3 is the final (or n2)

n=? is the initial and the start which is n1 in equation (or x in my expamle)
n=3 is the final which is n2 in equation

So if the formula is:

ΔE = R ( 1 / n2^2 - 1 / n2^2)

Then isn't it:
ΔE = R ( 1 / 3^2 - 1 / x^2)
ΔE = R ( 0.1111 - 1 / x^2)

11. Nov 17, 2015

### Staff: Mentor

Sorry for the confusion I created. The original equation you had is correct:
But considering that the change in energy ($\Delta E$) is for the atom, then going from $n_1 > n_2$ down to $n_2$, you have $\Delta E < 0$.

12. Nov 17, 2015

### ptownbro

Ok. Thanks. Just so I'm clear =D...

The answer is x = 2.27

Do you just round down to 2 and say that if went from n=2 to n=3?

13. Nov 17, 2015

### Staff: Mentor

This answer is not correct. First, it is too far from a whole number. Second, it has to be greater than 3 (look at the problem statement: the atom is initially in a higher energy level). Look at post #9 for the equality with the correct signs.

14. Nov 17, 2015

### ptownbro

Ok. I saw your edits and let's see if I've got this straight now.

Starting from my result for energy where I found E = 18.07*10^-20 J. If I've understood you, if the level is going up, then I use the positive version of this number. If it's falling I have to make it a negative. Therefore, the start of my formula should be (-18.07*10^-20 J) instead of (+18.07*10^-20 J) as in below (the right side of the equation was okay as I had it originally):

(-18.07*10^-20 J) = (-2.18*10^-18 J) * (1/3^2) - 1/x^2)

The rest would flow as:

(-18.07*10^-20 J) / (-2.18*10^-18 J) = (.1111 - 1/x^2)
(-0.1807*10^-18 J) / (-2.18*10^-18 J) = (0.1111 - 1/x^2)
(-0.1807) / (-2.18) = (0.1111 - 1/x^2)
0.0829 = 0.1111 - 1/x^2
0.0829 - 0.1111 = - 1/x^2
x^2 = -1 / -.0282
x = 35.448^(1/2)
x = 5.95

rounded x = 6

So the level is going from n=6 to n=3.

15. Nov 17, 2015

### Staff: Mentor

Correct. You can always check by plugging back the values in the original equation: find $\Delta E$ when $n_1 = 6$ and $n_2 = 3$.

16. Nov 17, 2015

### ptownbro

Yes! Thank you for all your help and patience.