anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@glutich @abdul_shabeer @Kitten_is_back @adajiamcneal @stonewoods @HazelLuv99 @Callisto
anonymous
  • anonymous
@gamer56 @KIT-KAT-KATE @Anon101 @JFraser
gamer56
  • gamer56
Hello Ella, First do you know any of the steps to completing this problem ?

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

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