Question

Three successive lines in the atomic spectrum of hydrogen have wave lenghts 18746A , 12815A and 10935A calculate the series limit for the spectral series to which the above lines belong?

Answer 1

@ghazi

Answer 2

@phi @mukushla @Callisto

Answer 3

i think you've to use Rydberg's formula

Answer 4

\[\frac{ 1 }{ \lambda }= R_{h}*(\frac { 1 }{ n _{1}^2 }-\frac{ 1 }{ n _{2}^2 })\]

Answer 5

8208 Amperes?

Answer 6

??

Answer 7

what's the answer?

Answer 8

@Yahoo!

Answer 9

@DLS u r correct

Answer 10

but it is no ampere it is Amstrong

Answer 11

ah sorry yeah armstrong

Answer 12

Can u show hw u got it

Answer 13

no :P

Answer 14

why

Answer 15

XD wher edid you try?

Answer 16

i have No idea..))

Answer 17

i nw Rydberg's formula....)) but i dont find a way to use it here

Answer 18

@ghazi any idea

Answer 19

hm....Can u show...@DLS

Answer 20

See you are given 3 successive wavelengths
Apply that formula :)

Answer 21

and we have to find n..))

Answer 22

you wil lget 3 equations

Answer 23

@ghazi the answer key tells like this 912* 9 = 8208 A

Answer 24

and wat does they really mean...

Answer 25

i think it's wavelength multiplied by 3^2 that is value of n..but i ain't sure

Answer 26

and wat is 912??

Answer 27

\[\frac{ 1 }{ 18746*10^{-10} }= R _{h}(\frac{ 1 }{ n _{1}^2 }-\frac{ 1 }{ n _{2}^2 })\] you'll find three equations by this and solve for it

Answer 28

lol....ok...912..))) rydberg consatnt ---- 109677

Answer 29

like i told you :)

Answer 30

yess...Rh= 1.0973*10^7 m

Answer 31

and wat is 912

Answer 32

a term :)