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jessi90

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what is the longest wavelength of the Balmer series?
 

Pwnage101

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well the Balmer series is:

let w= wavelength
n = prinicpal quantum number SQUARED, thus n>4 & n = 9, 16, 25, etc

1/w = R (1/4-1/n) = R((n-4)/4n)

since balmer series ends on n=2

thus w=(1/R)*((4n)/(n-4))

since 1/R is constant, w will be largest (longest) when 4n/(n-4) is largest, which occurs for the smallest possible value of n (by substiuting values), the smallest n = 9, thus the 3rd orbital shell

thus sub in and u get :

w=(1/R)*(36/5)
= 36/5R

R=1.097*10^7

w=6.56*10^-7m

yes, i know it can be done without the formal mathematics, but its good practice!!!
 
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umaycallmealex

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is Rh ever different? like if we are doing it with a different atom? or do we not have to/theorem inapplicable?
 

Pwnage101

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jessi90 said:
thanks, you are the pwnage!
glad to help one of the few ladies that do physics (couldnt believe it was 75% males and 25% females, was thinkin more 60/40, but apparently not)

good luk in the HSC!
 

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