Energy, wavelength

[needhelpperson]

The hydrogen atom can absorb light of wavelength 4055nm. Determine the inital and final values of n associated with this absorption.

Well i figured out the Energy it absorbed

(h*c)/(wavelength)

= ((6.63*10-34)*(3*10^8))/(4055/10^9)

=4.905 * 10^-20 J

so 4905*10^-20 = -2.18*10-18*(1/nf^2 - 1/ni^2)

1/nf^2 - 1/ni^2 = -.0225

How do i solve these two unknowns, other than guess and check?

[Sirus]

You know the energy associated with the first energy level of the hydrogen atom. You should also have a formula relating energy at nth level, energy at first level, and n. Did you not develop this formula in class?

[needhelpperson]

You know the energy associated with the first energy level of the hydrogen atom. You should also have a formula relating energy at nth level, energy at first level, and n. Did you not develop this formula in class?

no we did not. Please explain it. I would greatly appreciate it...

[Sirus]

The derivations are tedious, and although I always advise you to know where a formula came from before you use it, here it is:
E_n=\frac{E_1}{n^2}
Another useful one relates radius and energy level:
r_n=n^{2}r_1
Look these up in a textbook or on the internet to see where they come from.

[needhelpperson]

The derivations are tedious, and although I always advise you to know where a formula came from before you use it, here it is:
E_n=\frac{E_1}{n^2}
Another useful one relates radius and energy level:
r_n=n^{2}r_1
Look these up in a textbook or on the internet to see where they come from.


after looking at those formulas, How are they suppose to help me solve the two unkowns? How can I apply them? the inital n is not 1.

[Sirus]

I don't see why the initial n should be anything other than 1. It will only be higher if the atom has already had an electron bumped up to a higher energy level before absorbing the 4055-nm light. I'm not sure why they ask for initial n; it should just be 1.

[needhelpperson]

I don't see why the initial n should be anything other than 1. It will only be higher if the atom has already had an electron bumped up to a higher energy level before absorbing the 4055-nm light. I'm not sure why they ask for initial n; it should just be 1.

I already know the answer to this question. All i need is a more practical method to solve problems like these other than just using "guess and check".
As you were saying "It will only be higher if the atom has already had an electron bumped up to a higher energy level... ", that is what we're assuming.

If we were to assume initial n is 1

nf = root of 1/.9775

nf has to be a whole number and it is closest to 1. However this is not the answer, otherwise there would be no energy change. (1-1 = 0)

[Sirus]

I got n=6 using the energy formula from my earlier post. I'm not sure what you are doing above. Here is my work:

n=\sqrt{\frac{E_1}{E_n}}=\sqrt{\frac{13.6 eV}{\frac{4.905\times10^{-20}J}{\frac{1.60\times10^{-19}J}{eV}}}}=6.660547651
Therefore n=6 is the final energy level. Is that the answer?

[needhelpperson]

I got n=6 using the energy formula from my earlier post. I'm not sure what you are doing above. Here is my work:

n=\sqrt{\frac{E_1}{E_n}}=\sqrt{\frac{13.6 eV}{\frac{4.905\times10^{-20}J}{\frac{1.60\times10^{-19}J}{eV}}}}=6.660547651
Therefore n=6 is the final energy level. Is that the answer?

The formula n = E1/En is not the change in energy formula, which is what i needed. I think you mentioned before that it's just a ratio comparison. However though, the energy given in this question is the change in energy not the energy level at a particular level, for instance, 6.

Well lets test your nf = 6

We were given this Formula as the change in energy
planks constant * speed of light / the wavelength of the atom = -2.18 * 10^-18 * (1/nf^2-1/ni^2)

well the change in enery would be 4.905*10^-20 J

(-2.18*10^-18)*(1/36-1) = 2.12 *10^-18 J
Obviously 6 is not the right answer since it does not match 4.905*10^-20 J
And thus the initial n is not 1.


The answers are nf = 5 ni = 4. Try it.
But im still waiting to see if anyone has a more practical method.

[Sirus]

Well, your equation is the Energy eqaution equalled to the Rydberg equation. However, since the latter equation is equal to \frac{1}{\lambda}, the correct equation would be E=hc times the Rydberg eq. I got a weird answer with that though.

I still don't see why you don't think my method will work. I am finding the energy level associated with the energy of the photons of the light (not all of the energy is absorbed, because there is not enough to reach n=7 and more than enough for n=6).

[needhelpperson]

Well, your equation is the Energy eqaution equalled to the Rydberg equation. However, since the latter equation is equal to \frac{1}{\lambda}, the correct equation would be E=hc times the Rydberg eq. I got a weird answer with that though.

I still don't see why you don't think my method will work. I am finding the energy level associated with the energy of the photons of the light (not all of the energy is absorbed, because there is not enough to reach n=7 and more than enough for n=6).

Like i said before, the energy given is the change in energy. Not the energy at an orbital.

And i showed you using the Rydberg's equation for the change in energy that it didn't work in nf = 6.

[Sirus]

I guess your method is probably the best one then. I usually use my equations for this type of thing, but if your method works then you can use that.

This was fun :smile:

Sirus