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Please help I will medal! 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

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

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Other answers:

No, this question is unlike any I have seen before.
Ok Ella, I will help you. Please give me a second.
Thank you so much!
What you do is Use E=hf f=E/h
the energy of the emitted wave must match the energy \(difference\) between the two levels that it falls between
E=difference of 2 energy levels h=plank's constant f=frequency of emission
Is this making any sense so far ?
First, you need to find the energy of the 435nm wave
Yes it is. Alright.
Sorry the yes it is was to @gamer56
@Ella31224 Ok. Im sorry ella, But im having internet problems. Maybe @JFraser Can finish helping you ?
@JFraser If you could help me I would really appreciate it.
Can you find the energy (E) of the 435nm wave?
Would the equation be E=(6.63*10^(-34) J*S)(435)?
almost, you're forgetting that \(1nm = 1*10^{-9}m\), so you need to include the exponent
\(435nm = 435*10^{-9}m\)
wait, stop that
Stop what?
I didn't see you'd actually put up the wrong equation
What equation should I use?
it should be \(c = \lambda * \nu\), where c is the speed of light and \(\lambda\) is the wavelength
that will get you the \(frequency\), which is \(\nu\). That frequency gets plugged into \(E = h*\nu\)
Alright so V= (3.0*10^(8) M/J)/(435 x 10^-9 m)?
the units are m/s and m, but yes
that will get you the frequency. Plug that frequency into the energy equation and get the energy of the wave
that energy will be in \(joules\), so you need to use the eV-J conversion to find the energy in eV
Ok can you solve it with me just ensure I get the right answer?
What did you get, I'll check
the frequency should be a huge number, like \(10^{14}\)
yeah Im not getting a very large number..
if you're not careful, the order of operations on the calculator will give you a really random answer
Can you show me your order of operations because I am not getting a large number.
I got 0.6896551724137931 as my frequency
@Smita12 Could you help me?
:(
you need to make sure that the exponents get handled the right way. The conversion should look like\[\frac{3*10^8\frac{m}{s}}{435*10^{-9}m}\]
Was that the incorrect frequency?
that will \(get \space you\) the frequency, but you have to do out the math
Again I got 6.8965517e+14..
right "e+14" is the calculator's way of telling you scientific notation
so it is 6^14?
so you did get the right frequency
there's a big difference between \(6^{14}\), and \(6*10^{14}\)
Sorry 6 x 10^14
that's the frequency
Awesome! So now plug it into the energy equation?
E = (6.63 x 10^(-34) J*S)(6 x 10^14) correct?
correct
and you should get an energy that's pretty small
3 x 10^-19?
sounds about right
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
Alright
The closest eV to my answer is E3:E= -1.51eV.
What answer did you get?
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
Alright so how would I go about doing this?
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
that makes the eV difference about 2.85eV
Alright and is that the correct answer?
again, it's the \(difference\) between two levels that must equal about 2.85 eV
the energy levels sort of look like this|dw:1432832938799:dw|
|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
So the two levels would be E3 and E2?
3.85eV - 1.5eV is only about 2.3eV, can there be a better choice?
E4 and E2?
that's only 2.6 (I like E5 - E2)
Awesome thank you! So the answer to my question is E5 - E2?
that's what it seems to me
Thank you so much! You helped me learn a lot!
YW

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