Converting Energy Lines from Electron Volts

In summary, the energy line for an electron dropping from one energy level to another in the energy diagram for hydrogen would show a line at 435 nm in an emission spectrum.
  • #1
Illuminitwit
22
0

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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  • #2
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...
 
  • #3
So in the graph, the -.04 eV is the ∆eV? eV(before dropping energy level) - eV(after falling energy level) = ∆eV?
 
  • #4
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...
 
  • #5
Whew! Thanks so much, I think that'll help a LOT! :)
 
  • #6
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?
 
  • #7
∆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?
 
  • #8
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.
 
  • #9
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.
 

Related to Converting Energy Lines from Electron Volts

1. What is the concept of converting energy lines from electron volts?

The concept of converting energy lines from electron volts is based on the idea of converting energy units from electron volts (eV) to other units such as joules (J). This conversion is necessary when working with different energy scales and is commonly used in the field of physics and chemistry.

2. Why is it important to convert energy lines from electron volts?

Converting energy lines from electron volts is important because it allows for better understanding and comparison of different energy values. Since electron volts are commonly used in atomic and subatomic physics, converting them to other units such as joules allows for easier analysis and calculation of energy quantities.

3. How do you convert energy lines from electron volts to joules?

To convert energy lines from electron volts to joules, you can use the conversion factor 1 eV = 1.602 x 10^-19 J. Simply multiply the energy value in eV by this conversion factor to get the equivalent value in joules.

4. Can you convert energy lines from electron volts to other units besides joules?

Yes, you can convert energy lines from electron volts to other units such as ergs, calories, or kilocalories. The conversion factor will vary depending on the unit you are converting to, but the process is the same – multiply the energy value in eV by the appropriate conversion factor.

5. Is there a shortcut to convert energy lines from electron volts to other units?

Yes, there are online calculators and conversion charts available that can help you quickly and accurately convert energy lines from electron volts to other units. However, it is still important to understand the concept and the conversion factor in order to ensure accuracy and avoid errors in calculations.

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