How Do You Determine Quantum Numbers from Hydrogen Atom Wavelengths?

[Ush]

Homework Statement

Two of the wavelengths emitted by a hydrogen atom are 9.50×10-8 m and 4.65×10-6 m. What are the m and n values for the first wavelength?

Homework Equations

E = -13.6eV/n²
E = -13.6eV/(m² - n²)
restriction:m > n
E = hf
f = c/λ
1 eV = 1.6 x 10^-19J

The Attempt at a Solution

E = -13.6eV/(m² - n²)
hc/λ = -13.6eV/(m² - n²)
hc/λ(-13.6eV) = 1/(m² - n²)
λ(-13.6eV)/hc = m² - n²
λ(2.179E-18J)/hc = m² - n²

let λ =9.50×10-8

(9.50×10-8)(2.179E-18J)/hc = m² - n²
1 = m² - n²
1 = (m-n)(m+n)

...now what? o_o
...am I doing this right?
...if I do the same process with the other wavelength then I get

51 = m_b² - n_b²
51 = (m_b-n_b)(m_b+n_b)

...but nothing says m and m_b or n / n_b are related to each other..

this looks like something I did in h.s. -I don't remember how I did it =[

please help

 

[hikaru1221]

It should be: E = -13.6 ev * (1/n² - 1/m²)

 

[kuruman]

Homework Statement

Two of the wavelengths emitted by a hydrogen atom are 9.50×10-8 m and 4.65×10-6 m. What are the m and n values for the first wavelength?

Homework Equations

E = -13.6eV/n²
E = -13.6eV/(m² - n²)
restriction:m > n

Incorrect. Should be E = -13.6 eV {(1/n²)-(1/m²)}!

1/4 - 1/3 is NOT 1/(4-3) = 1/1 !

 

[Ush]

what's the difference?
if m > n then.. ex.. if m = 3 and n = 2 ... 1/3^2 - 1/2^2 = a negative number! a negative number multiplied by -13.6 is a positive number :S

 

[Ush]

i mean, if I switch the two, won't I have to switch the restriction anyway? to make E positive? I'm confused =S

 

[graphene]

read carefully, what hikaru & kuru are pointing out.

 

[Ush]

Oh,

-13.6eV/(m² - n²) ≠ E = -13.6 ev * (1/n² - 1/m²)

but if m > n, then shouldn't it be
E = -13.6 ev * (1/m² - 1/n²)
in order to get a positive energy? =S

if I put n before m i'll get a negative energy,
that doesn't make sense because photons have to have a positive energy!
I'm almost certain of that =/

Try 2:

E = -13.6 ev * (1/m² - 1/n²)
hc/λ = -13.6 ev * (1/m² - 1/n²)
hc/λ(-13.6eV) = (1/m² - 1/n²)

let λ =9.50×10-8

-0.96 = (1/m² - 1/n²)
and now..?

 

[kuruman]

Oh,

-13.6eV/(m² - n²) ≠ E = -13.6 ev * (1/n² - 1/m²)

Correct, now you see the difference.

but if m > n, then shouldn't it be
E = -13.6 ev * (1/m² - 1/n²)
in order to get a positive energy? =S

if I put n before m i'll get a negative energy,
that doesn't make sense because photons have to have a positive energy!
I'm almost certain of that =/

The formula that you are using gives the change in energy of the atom. If it is positive, the atom gains energy, if it is negative it loses energy. And yes, a photon always has positive energy. You should be 100% sure of that. Take the absolute value of the difference as the energy of the photon, which could either be gained or lost by the atom.

Try 2:

E = -13.6 ev * (1/m² - 1/n²)
hc/λ = -13.6 ev * (1/m² - 1/n²)
hc/λ(-13.6eV) = (1/m² - 1/n²)

let λ =9.50×10-8

-0.96 = (1/m² - 1/n²)
and now..?

Try different pairs of integers m and n, but be systematic about it.

 

[Ush]

using different pairs of integers..
1/5² - 1/1 = -0.96

is there no mathematical way of doing this, other then guessing?

 

[kuruman]

using different pairs of integers..
1/5² - 1/1 = -0.96

is there no mathematical way of doing this, other then guessing?

Think about it. You have one equation and two unknowns.

Of course, you can fix n and then find the shortest wavelength of a photon (highest energy) that is emitted by a transition to that level. This means that m is infinite, or 1/m² = 0. In that case, the shortest wavelength to the n= 2 level is 2.29x10^-6 m, which is much longer than 9.6x10^-8 m. Any n other than 1 will give you even longer wavelengths. Therefore n must be equal to 1. Once you establish that, there is a finite number of m values that will do the job. That is what I mean by "systematic".

 

[Ush]

oh..
so giving us another wavelength in the question is just to confuse us right?

 

[Ush]

thank you for your help =)

 

[kuruman]

oh..
so giving us another wavelength in the question is just to confuse us right?

I don't know, and I cannot read the mind of the person who wrote the problem. Anyway, I think you now understand how this works.