Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

Thank you so much!

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Is this making any sense so far ?

Answer 11

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

Answer 12

Yes it is. Alright.

Answer 13

Sorry the yes it is was to @gamer56

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

wait, stop that

Answer 21

Stop what?

Answer 22

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

Answer 23

What equation should I use?

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

the units are m/s and m, but yes

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

What did you get, I'll check

Answer 32

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

Answer 33

yeah Im not getting a very large number..

Answer 34

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

Answer 35

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

Answer 36

I got 0.6896551724137931 as my frequency

Answer 37

@gamer56

Answer 38

@Smita12 Could you help me?

Answer 39

:(

Answer 40

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

Answer 41

Was that the incorrect frequency?

Answer 42

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

Answer 43

Again I got 6.8965517e+14..

Answer 44

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

Answer 45

so it is 6^14?

Answer 46

so you did get the right frequency

Answer 47

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

Answer 48

Sorry 6 x 10^14

Answer 49

that's the frequency

Answer 50

Awesome! So now plug it into the energy equation?

Answer 51

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

Answer 52

correct

Answer 53

and you should get an energy that's pretty small

Answer 54

3 x 10^-19?

Answer 55

sounds about right

Answer 56

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

Answer 57

Alright

Answer 58

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

Answer 59

What answer did you get?

Answer 60

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

Answer 61

Alright so how would I go about doing this?

Answer 62

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

Answer 63

that makes the eV difference about 2.85eV

Answer 64

Alright and is that the correct answer?

Answer 65

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

Answer 66

the energy levels sort of look like this

Answer 67

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