Question

could someone clarify this question?

Answer 1

we did a lab and in this lab I recored the four lines wavelengths as "409.00nm", "433.00","488.00" & "655.00"

Answer 2

Do you know how to use the Rydberg equation?

Answer 3

@aaronq is it this equation "- RH [ (1/(nf)^2) - (1/(ni)^2) ] "

Answer 4

I'm just unsure how will i incorporate the wavelengths I recorded to find ni ?

Answer 5

you're missing part of it, \(\dfrac{1}{\lambda}=RZ(\dfrac{1}{n_i^2}-\dfrac{1}{n_f^2})\)

Answer 6

so you just need to isolate \(n_i\)

Answer 7

@aaronq ahhhh okay i understand. in my lab manual the equation is a little different instead of (1/λ) it is (hc/λ) ?

Answer 8

as for "nf" that will always be 2 correct?

Answer 9

There are lot of variations depending on what value for R you use :S

You can also use that one, h and c are just constants.

i would think that \(n_f\) would equal 1, since that's the lowest level it can fall to, but you might have to try a few numbers if you don't get integers.

Answer 10

@aaronq oh ok i just realized they were constants as well! hm ok so would you suggest setting "nf" as 1?

Answer 11

yeah most definitely, but if you end up with non-integers, try 2.

Answer 12

@aaronq ok! i actually tried it out and i ended up with ni=0.904 :S

Answer 13

you mean \(n_f\)=0.904?

you should try it with \(n_i\)=2, if it's way off, then go with \(n_i\)=1.

Answer 14

no ni, because the question asks to find "ni" :S

Answer 15

@aaronq

Answer 16

hm i'll try it out, which wavelength did you use?

Answer 17

thank you :/ my first one was "409.00"

Answer 18

and you used \(\dfrac{hc}{\lambda}\), what's your R constant?

Answer 19

i set nf = 1 & i got 0.904. but i just don't know if thats what I'm supposed to be doing ?

Answer 20

no, i used the equation you provided. & for R i used "0.01097"

Answer 21

which is rydbergs constant

Answer 22

yeah, that shouldn't happen, let me check if you made an error in the math.

Answer 23

ok! i really appreciate your help, as always!

Answer 24

hmm if you use the one i posted you need to use R= \(1.097*10^7 ~/m\)

Answer 25

isn't it "-7" ?

Answer 26

nope, they must have taken other constants into account for get the -7 exponent.

Oh i just noticed i gave you the formula incorrectly, it should be:

\(\dfrac{1}{\lambda}=1.097*10^7(\dfrac{1}{n^2_f}-\dfrac{1}{n_i^2})\)

so if i put \(n_f\)=2, i get \(n_i\)=6

Answer 27

ah! i noticed i made a couple mistakes :| but ok i will try to do it again and see if i get the same :)

Answer 28

i'm pretty sure you were doing it right, i think i got the same answer as you when i had the n's in the wrong places.

Answer 29

@aaronq hm i dont understand for some reason I'm not getting the same answer as you

Answer 30

did you have: \(\dfrac{1}{(409*10^{-9}m)}=(1.097*10^7/m)(\dfrac{1}{2^2}-\dfrac{1}{x^2})\)

Answer 31

i got it! :D ah simple mistakes :/ but not i think i am cable of doing the rest. thank you so much!

Answer 32

i know, it's so easy to make mistakes because the equation is so big. No problem ! good luck :)

Answer 33

@aaronq thank you! i was able to solve for the rest of the wavelengths and i think i did good :) HOWEVER, I'm afraid i have one more question!

Answer 34

great! haha not a problem, shoot.

Answer 35

hehe ok, well the next question is as follows :

Answer 36

lets say i were to calculate ni for Helium would i set "nf" = 2 or would it be different?

Answer 37

you would have to change the Rydberg constant because the number of protons, Z, will be 2 for He.

Answer 38

hm, and how would i change the constant? not sure how that is possible

Answer 39

\(\dfrac{1}{\lambda}=RZ(\dfrac{1}{n_f^2}...etc)\)


Z is the number of protons, so for He, Z=2

Answer 40

oh ok i see what you did there! & "nf" would again be set to = 2?

Answer 41

or would that have to change as well?

Answer 42

oh wait sorry, it's \(Z^2\)

hm idk, try it as 2 first, if not then try other numbers.

Answer 43

lol. ok im a tad confused so are you saying this would be my set up ...\[\frac{ 1 }{ \lambda }= (1.097\times10^{7})^{2}(\frac{ 1 }{ nf ^{2} }..\]

Answer 44

close, \(\dfrac{1}{\lambda}=(1.097×10^7/m)Z^2(.....)\)

Z=2, so \(\dfrac{1}{\lambda}=(1.097×10^7/m)2^2(.....)\)

Answer 45

ooo i gotcha :) can't thank you enough! for your help and patience

Answer 46

sweet! no problem, really :) I hope you're gotta questions though because i gotta go lol