Atomic absorption spectrum

  • #1
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Homework Statement



The following photon wavelengths are observed in absorption at room temperature from an ionized atomic gas with a single electron orbiting the nucleus: λ=
13:5 nm, 11:4 nm, 10:8 nm. Use this data to determine the effective Rydberg constant
and the nuclear charge.

Homework Equations





The Attempt at a Solution



I know that energy is emitted/absorbed when electron transits between different energy levels.
2hxc0ag.png


Thing is, I don't even know what the n=1 energy is, to find the rydberg constant. I can't tell what the energy levels are just from the photons absorbed, as they could be between any two levels.
 

Answers and Replies

  • #2
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11,862
Don't give up before you even started.
Did you convert the wavelength values to photon energies?
Which patterns do you expect in the energy spectrum?
 
  • #3
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Don't give up before you even started.
Did you convert the wavelength values to photon energies?
Which patterns do you expect in the energy spectrum?

I expect, the gaps between energy levels to become smaller as you go higher up..
 
  • #4
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13
Don't give up before you even started.
Did you convert the wavelength values to photon energies?
Which patterns do you expect in the energy spectrum?

13.5nm -> 92.1 eV
11.4nm -> 109.0 eV
10.8nm -> 115.1 eV

So

n=1 is 92.1 eV
n=2 is 109.0 eV
n=3 is 115.5 eV

Then what is the point of giving us the other 2 absorption wavelengths?
 
  • #5
35,437
11,862
So

n=1 is 92.1 eV
n=2 is 109.0 eV
n=3 is 115.5 eV

Then what is the point of giving us the other 2 absorption wavelengths?
Photon energies are not the energies of states!

Which photon energies (not electron energy states) do you get for a hydrogen atom? Which ratios do you have between those values? Do you see a similarity to your problem?
 
  • #6
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For Hydrogen:

E1 = -13.6 eV
E2 = -3.4 eV
E3 = -1.51 eV
E4 = -0.850 eV

Based on the wavelengths given, 92.1eV, 109eV, 115.5eV are differences in energy between En and E1.

The ratio of first energy level between the gas and hydrogen = Z2, where Z is the proton number of the gas.
 
  • #7
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Photon energies are not the energies of states!

Which photon energies (not electron energy states) do you get for a hydrogen atom? Which ratios do you have between those values? Do you see a similarity to your problem?

Ok using the relation ΔE = (1- 1/n2)

First emission:
92.1 = E1(3/4)

Second emission:
109 = E1(8/9)

Third Emission:

115.5 = E1(15/16)

These ratios match, so somehow these are the emissions from the second, third and fourth energy levels.

E1 = (13.6)Z2

Solving, Z = 3 (Lithium) and E1 = 122.4eV, R = (13.6eV)/hc = 1.09*107
 
Last edited:
  • #8
35,437
11,862
That is correct.

R = (13.6eV)/hc = 1.09*107
I think the factor of 9 is missing here, and the last value should have units.
 

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