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I am confused on how to solve this question:
Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one energy level to another would show a line at 435 nm in an emission spectrum. Use the formulas and the energy level information below to answer the question.
Needed constants:
1.00eV=1.6*10^(-19) J
C= 3.0*10^(8) M/J
H= 6.63*10^(-34) J*S
Energy Level Values:
E6:E= -0.378eV
E5:E= -0.544eV
E4:E= -0.850eV
E3:E= -1.51eV
E2:E= -3.403V

- anonymous

- schrodinger

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- anonymous

@glutich @abdul_shabeer @Kitten_is_back @adajiamcneal @stonewoods @HazelLuv99 @Callisto

- anonymous

@gamer56 @KIT-KAT-KATE @Anon101 @JFraser

- gamer56

Hello Ella, First do you know any of the steps to completing this problem ?

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## More answers

- anonymous

No, this question is unlike any I have seen before.

- gamer56

Ok Ella, I will help you. Please give me a second.

- anonymous

Thank you so much!

- gamer56

What you do is Use
E=hf
f=E/h

- JFraser

the energy of the emitted wave must match the energy \(difference\) between the two levels that it falls between

- gamer56

E=difference of 2 energy levels
h=plank's constant
f=frequency of emission

- gamer56

Is this making any sense so far ?

- JFraser

First, you need to find the energy of the 435nm wave

- anonymous

Yes it is. Alright.

- anonymous

Sorry the yes it is was to @gamer56

- gamer56

@Ella31224 Ok. Im sorry ella, But im having internet problems. Maybe @JFraser Can finish helping you ?

- anonymous

@JFraser If you could help me I would really appreciate it.

- JFraser

Can you find the energy (E) of the 435nm wave?

- anonymous

Would the equation be E=(6.63*10^(-34) J*S)(435)?

- JFraser

almost, you're forgetting that \(1nm = 1*10^{-9}m\), so you need to include the exponent

- JFraser

\(435nm = 435*10^{-9}m\)

- JFraser

wait, stop that

- anonymous

Stop what?

- JFraser

I didn't see you'd actually put up the wrong equation

- anonymous

What equation should I use?

- JFraser

it should be \(c = \lambda * \nu\), where c is the speed of light and \(\lambda\) is the wavelength

- JFraser

that will get you the \(frequency\), which is \(\nu\). That frequency gets plugged into \(E = h*\nu\)

- anonymous

Alright so V= (3.0*10^(8) M/J)/(435 x 10^-9 m)?

- JFraser

the units are m/s and m, but yes

- JFraser

that will get you the frequency. Plug that frequency into the energy equation and get the energy of the wave

- JFraser

that energy will be in \(joules\), so you need to use the eV-J conversion to find the energy in eV

- anonymous

Ok can you solve it with me just ensure I get the right answer?

- JFraser

What did you get, I'll check

- JFraser

the frequency should be a huge number, like \(10^{14}\)

- anonymous

yeah Im not getting a very large number..

- JFraser

if you're not careful, the order of operations on the calculator will give you a really random answer

- anonymous

Can you show me your order of operations because I am not getting a large number.

- anonymous

I got 0.6896551724137931 as my frequency

- anonymous

@gamer56

- anonymous

@Smita12 Could you help me?

- anonymous

:(

- JFraser

you need to make sure that the exponents get handled the right way. The conversion should look like\[\frac{3*10^8\frac{m}{s}}{435*10^{-9}m}\]

- anonymous

Was that the incorrect frequency?

- JFraser

that will \(get \space you\) the frequency, but you have to do out the math

- anonymous

Again I got 6.8965517e+14..

- JFraser

right "e+14" is the calculator's way of telling you scientific notation

- anonymous

so it is 6^14?

- JFraser

so you did get the right frequency

- JFraser

there's a big difference between \(6^{14}\), and \(6*10^{14}\)

- anonymous

Sorry 6 x 10^14

- JFraser

that's the frequency

- anonymous

Awesome! So now plug it into the energy equation?

- anonymous

E = (6.63 x 10^(-34) J*S)(6 x 10^14) correct?

- JFraser

correct

- JFraser

and you should get an energy that's pretty small

- anonymous

3 x 10^-19?

- JFraser

sounds about right

- JFraser

that's in \(joules\), but the energy levels you're given are in eV. so use the conversion to find the energy of that wave in eV

- anonymous

Alright

- anonymous

The closest eV to my answer is E3:E= -1.51eV.

- anonymous

What answer did you get?

- JFraser

it's not the energy level that's important, because the wave is created by the electron moving \(between\) two levels. You need to find two levels that have a \(difference\) of 1.5eV

- anonymous

Alright so how would I go about doing this?

- JFraser

when you calculated the frequency, you cut off a few too many digits. the frequency should be closer to 6.8*10^14Hz, which makes the energy closer to 4.6*10^-19J

- JFraser

that makes the eV difference about 2.85eV

- anonymous

Alright and is that the correct answer?

- JFraser

again, it's the \(difference\) between two levels that must equal about 2.85 eV

- JFraser

the energy levels sort of look like this|dw:1432832938799:dw|

- JFraser

|dw:1432833072141:dw|
when the wave is created, the energy \(difference\) between two levels is equal to the energy of the wave.
You know the energy of the wave is about 2.85eV, so find two levels separated by about 2.85eV

- anonymous

So the two levels would be E3 and E2?

- JFraser

3.85eV - 1.5eV is only about 2.3eV, can there be a better choice?

- anonymous

E4 and E2?

- JFraser

that's only 2.6
(I like E5 - E2)

- anonymous

Awesome thank you! So the answer to my question is E5 - E2?

- JFraser

that's what it seems to me

- anonymous

Thank you so much! You helped me learn a lot!

- JFraser

YW

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