- #1

Illuminitwit

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## Homework Statement

"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 10

^{8}m/s

E

_{n}= 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 10

^{8}/ 4.35 x 10

^{-7}= 6.87 x 10

^{14}

6.87 x 10

^{14}•*h = E

6.87 x 10

^{14}• 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 10

^{14}• h • n = E

6.87 x 10

^{14}• 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"?