Converting Energy Lines from Electron Volts

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SUMMARY

The forum discussion focuses on converting energy lines from electron volts (eV) to wavelengths in nanometers (nm) for hydrogen's emission spectrum. The key equations used include E = hf, λ = c/f, and the conversion factor 1 eV = 1.6 x 10-19 J. The participant successfully calculated the energy level transition corresponding to a wavelength of 435 nm, identifying that the transition occurs from E2 to E5 with a change of approximately 2.9 eV. The discussion emphasizes the importance of calculating energy deltas rather than using absolute energy values.

PREREQUISITES
  • Understanding of quantum mechanics and energy level transitions
  • Familiarity with the equations E = hf and λ = c/f
  • Knowledge of unit conversions between electron volts and joules
  • Basic proficiency in algebra for manipulating equations
NEXT STEPS
  • Study the concept of energy level transitions in hydrogen and other elements
  • Learn about the significance of emission spectra in spectroscopy
  • Explore the derivation and applications of the Planck-Einstein relation E = hf
  • Investigate common mistakes in unit conversions and dimensional analysis
USEFUL FOR

Students studying quantum mechanics, physics educators, and anyone interested in understanding atomic emission spectra and energy transitions in hydrogen.

Illuminitwit
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Homework Statement



Henergy.GIF


"Using the conversion factor from eV (electron volts) to Joules, determine which energy line for an electron dropping from one energy level to another in the energy diagram for hydrogen above would show a line at 435 nm in an emission spectrum."


Homework Equations


1 eV = 1.6 x 10-19 J
h = 6.63 x 10-34 J•s
c = 2.99 x 108 m/s
En = nhf
λ = c/f
λ = E/h
f = E/h


The Attempt at a Solution



First, I converted all the eV;
-.04 • 1.6 x 10-19 = -6.4 x 10-21
-.05 • 1.6 x 10-19 = -8 x 10-21
-.08 • 1.6 x 10-19 = -1.28 x 10-20
-1.5 • 1.6 x 10-19 = -2.4 x 10-19
-3.4 • 1.6 x 10-19 = -5.44 x 10-19
-13.6 • 1.6 x 10-19 = -2.18 x 10-18

Next, I think I'm supposed to find all the frequencies using;
f = E/h

Or maybe I'm supposed to find the wavelengths with;
λ = c/f
λ = E/h (is this formula even correct?)

Could I just put it the desired answer and see which matches?;
4.35 x 10-7 = c/f
4.35 x 10-7 = E/h
f = E/h

c/ 4.35 x 10-7 = f
2.99 x 108 / 4.35 x 10-7 = 6.87 x 1014

6.87 x 1014 •*h = E
6.87 x 1014 • 6.63 x 10-34 = 4.55 x 10-19

It doesn't seem to match anything...

Retry previous steps adding n into the equation;

6.87 x 1014 • h • n = E
6.87 x 1014 • 6.63 x 10-34 = E/n
4.55 x 10-19 = E/n
4.55 x 10-19 • n = E

Go back to all n from original graph;

4.55 x 10-19 • 6 = 2.73 x 10-18
4.55 x 10-19 • 5 = 2.28 x 10-18
4.55 x 10-19 • 4 = 1.82 x 10-18
4.55 x 10-19 • 3 = 1.37 x 10-18
4.55 x 10-19 • 2 = 9.1 x 10-19
4.55 x 10-19 • 1 = 4.55 x 10-19

I don't exactly understand what I'm supposed to get for 435 nm. The "nm" means nanometers, right? Is that the wavelength? Is it the frequency? Is it something else...? What is "a line at 435 nm in an emission spectrum"?
 
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Illuminitwit said:

Homework Statement



Henergy.GIF


"Using the conversion factor from eV (electron volts) to Joules, determine which energy line for an electron dropping from one energy level to another in the energy diagram for hydrogen above would show a line at 435 nm in an emission spectrum."


Homework Equations


1 eV = 1.6 x 10-19 J
h = 6.63 x 10-34 J•s
c = 2.99 x 108 m/s
En = nhf
λ = c/f
λ = E/h
f = E/h


The Attempt at a Solution



First, I converted all the eV;
-.04 • 1.6 x 10-19 = -6.4 x 10-21
-.05 • 1.6 x 10-19 = -8 x 10-21
-.08 • 1.6 x 10-19 = -1.28 x 10-20
-1.5 • 1.6 x 10-19 = -2.4 x 10-19
-3.4 • 1.6 x 10-19 = -5.44 x 10-19
-13.6 • 1.6 x 10-19 = -2.18 x 10-18

Next, I think I'm supposed to find all the frequencies using;
f = E/h

Or maybe I'm supposed to find the wavelengths with;
λ = c/f
λ = E/h (is this formula even correct?)

Could I just put it the desired answer and see which matches?;
4.35 x 10-7 = c/f
4.35 x 10-7 = E/h
f = E/h

c/ 4.35 x 10-7 = f
2.99 x 108 / 4.35 x 10-7 = 6.87 x 1014

6.87 x 1014 •*h = E
6.87 x 1014 • 6.63 x 10-34 = 4.55 x 10-19

It doesn't seem to match anything...

Retry previous steps adding n into the equation;

6.87 x 1014 • h • n = E
6.87 x 1014 • 6.63 x 10-34 = E/n
4.55 x 10-19 = E/n
4.55 x 10-19 • n = E

Go back to all n from original graph;

4.55 x 10-19 • 6 = 2.73 x 10-18
4.55 x 10-19 • 5 = 2.28 x 10-18
4.55 x 10-19 • 4 = 1.82 x 10-18
4.55 x 10-19 • 3 = 1.37 x 10-18
4.55 x 10-19 • 2 = 9.1 x 10-19
4.55 x 10-19 • 1 = 4.55 x 10-19

I don't exactly understand what I'm supposed to get for 435 nm. The "nm" means nanometers, right? Is that the wavelength? Is it the frequency? Is it something else...? What is "a line at 435 nm in an emission spectrum"?

Remember that the transition of the electron is *between* those energy levels in the diagram, so you should be calculating the deltas, not using the energies themselves. And E=hf is the correct equation to use after that...
 
So in the graph, the -.04 eV is the ∆eV? eV(before dropping energy level) - eV(after falling energy level) = ∆eV?
 
Illuminitwit said:
So in the graph, the -.04 eV is the ∆eV? eV(before dropping energy level) - eV(after falling energy level) = ∆eV?

The energy delta for an n = 6-->5 transition is 0.1eV, and so on. You can also try working backwards -- that may be quicker. Calculate the photon energy in eV for the wavelength of light that you are given, and look for a delta-eV transision that matches on the graph...
 
Whew! Thanks so much, I think that'll help a LOT! :)
 
4.35 x 10-7 = c/f
c / 4.35 x 10-7 = f
2.99 x 108 / 4.35 x 10-7 = f
f = 6.87 x 1014

f = E/h
f • h = E
6.87 x 1014 • 6.63 x 10-34 = E
E = 1.04 x 1048

1.6 x 10-19 / 1.04 x 1048 = ∆eV

Oops... I'm sure I did something wrong... Do I at least have the right sort of idea? Or is my approach completely off?
 
∆eV = -.4 - (-.5) = .1 eV
f = E/h
f = (.1)(1.6 x 10-19) / 6.63 x 10-34
f = 2.4 x 1013
λ = c/f
λ = 2.99 x 108 / 2.4 x 1013
λ = 1.24 x 10-5

Working backwards;
λ = 4.35 x 10-7
4.35 x 10-7 = 2.99 x 108 / f
2.99 x 108 / 4.35 x 10-7 = f
f = 6.89 x 1014
f • h = E
f • 6.63 x 10-34 = ∆eV • 1.6 x 10-19

6.89 x 1014 • 6.63 x 10-34 / 1.6 x 10-19 = ∆eV
∆eV = 2.85 ≈ 2.9

So the energy level change is from E2 to E5 because -.5 - (-3.4) = 2.9 eV. Is that right?
 
Illuminitwit said:
∆eV = -.4 - (-.5) = .1 eV
f = E/h
f = (.1)(1.6 x 10-19) / 6.63 x 10-34
f = 2.4 x 1013
λ = c/f
λ = 2.99 x 108 / 2.4 x 1013
λ = 1.24 x 10-5

Working backwards;
λ = 4.35 x 10-7
4.35 x 10-7 = 2.99 x 108 / f
2.99 x 108 / 4.35 x 10-7 = f
f = 6.89 x 1014
f • h = E
f • 6.63 x 10-34 = ∆eV • 1.6 x 10-19

6.89 x 1014 • 6.63 x 10-34 / 1.6 x 10-19 = ∆eV
∆eV = 2.85 ≈ 2.9

So the energy level change is from E2 to E5 because -.5 - (-3.4) = 2.9 eV. Is that right?

I didn't check it in detail, but it looks correct. One thing that would help would be to include units on each line of your calculation, and carry them along with the quantities. That makes it a lot easier to follow the calculation, and helps you to catch errors in terms as you go.

Good job.
 
Good tip, and thanks for the help!

Also, I noticed some of my mistakes from the original equations. The change in eV, as you said, was .1, whereas I had the numbers displaced by a factor of 10 with an extra zero to the right of the decimal. It kind of screwed up my powers...

I think I finally have it, though, hehehe.
 

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