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The first emission line of Balmer series...

The first emission line of Balmer series in H spectrum has the wave number equal to :

A

`(9R_H)/(400) cm^(-1)`

B

`(9R_H)/(144)cm^(-1)`

C

`(3R_H)/4cm^(-1)`

D

`(5R_H)/(36) cm^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
`overlinev = 1/(lambda)= R_H [ 1/2^2 - 1/3^2], n_1 = 2` for Balmer series ,
` n_2 = 3` for `H_ (alpha)` line .
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