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anonymous
 one year ago
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
anonymous
 one year ago
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@glutich @abdul_shabeer @Kitten_is_back @adajiamcneal @stonewoods @HazelLuv99 @Callisto

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@gamer56 @KITKATKATE @Anon101 @JFraser

Gamer56
 one year ago
Best ResponseYou've already chosen the best response.1Hello Ella, First do you know any of the steps to completing this problem ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, this question is unlike any I have seen before.

Gamer56
 one year ago
Best ResponseYou've already chosen the best response.1Ok Ella, I will help you. Please give me a second.

Gamer56
 one year ago
Best ResponseYou've already chosen the best response.1What you do is Use E=hf f=E/h

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the energy of the emitted wave must match the energy \(difference\) between the two levels that it falls between

Gamer56
 one year ago
Best ResponseYou've already chosen the best response.1E=difference of 2 energy levels h=plank's constant f=frequency of emission

Gamer56
 one year ago
Best ResponseYou've already chosen the best response.1Is this making any sense so far ?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2First, you need to find the energy of the 435nm wave

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry the yes it is was to @gamer56

Gamer56
 one year ago
Best ResponseYou've already chosen the best response.1@Ella31224 Ok. Im sorry ella, But im having internet problems. Maybe @JFraser Can finish helping you ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@JFraser If you could help me I would really appreciate it.

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2Can you find the energy (E) of the 435nm wave?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Would the equation be E=(6.63*10^(34) J*S)(435)?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2almost, you're forgetting that \(1nm = 1*10^{9}m\), so you need to include the exponent

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2\(435nm = 435*10^{9}m\)

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2I didn't see you'd actually put up the wrong equation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What equation should I use?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2it should be \(c = \lambda * \nu\), where c is the speed of light and \(\lambda\) is the wavelength

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2that will get you the \(frequency\), which is \(\nu\). That frequency gets plugged into \(E = h*\nu\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright so V= (3.0*10^(8) M/J)/(435 x 10^9 m)?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the units are m/s and m, but yes

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2that will get you the frequency. Plug that frequency into the energy equation and get the energy of the wave

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2that energy will be in \(joules\), so you need to use the eVJ conversion to find the energy in eV

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok can you solve it with me just ensure I get the right answer?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2What did you get, I'll check

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the frequency should be a huge number, like \(10^{14}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah Im not getting a very large number..

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2if you're not careful, the order of operations on the calculator will give you a really random answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you show me your order of operations because I am not getting a large number.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got 0.6896551724137931 as my frequency

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Smita12 Could you help me?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2you 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
 one year ago
Best ResponseYou've already chosen the best response.0Was that the incorrect frequency?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2that will \(get \space you\) the frequency, but you have to do out the math

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Again I got 6.8965517e+14..

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2right "e+14" is the calculator's way of telling you scientific notation

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2so you did get the right frequency

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2there's a big difference between \(6^{14}\), and \(6*10^{14}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome! So now plug it into the energy equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0E = (6.63 x 10^(34) J*S)(6 x 10^14) correct?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2and you should get an energy that's pretty small

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2that'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
 one year ago
Best ResponseYou've already chosen the best response.0The closest eV to my answer is E3:E= 1.51eV.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What answer did you get?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2it'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
 one year ago
Best ResponseYou've already chosen the best response.0Alright so how would I go about doing this?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2when 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
 one year ago
Best ResponseYou've already chosen the best response.2that makes the eV difference about 2.85eV

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright and is that the correct answer?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2again, it's the \(difference\) between two levels that must equal about 2.85 eV

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2the energy levels sort of look like thisdw:1432832938799:dw

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2dw: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
 one year ago
Best ResponseYou've already chosen the best response.0So the two levels would be E3 and E2?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.23.85eV  1.5eV is only about 2.3eV, can there be a better choice?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2that's only 2.6 (I like E5  E2)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Awesome thank you! So the answer to my question is E5  E2?

JFraser
 one year ago
Best ResponseYou've already chosen the best response.2that's what it seems to me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! You helped me learn a lot!
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