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Question
Calculate the wavelength of the first two lines in the Balmer series of hydrogen atoms.
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Solution
For Balmer series,
n = 2, m = 3, 4, 5, ....
(a) For first line in Balmer series, n = 2, m = 3
`1/lambda_1 = R_H(1/n^2 - 1/m^2)`
= `R_H (1/(2)^2 - 1/(3)^3)`
= `R_H (1/4 - 1/9)`
∴ `1/lambda_1 = R_H5/36`
∴ `lambda_1 = 36/(5 R_H)`
Putting RH = 1.097 × 107 m−1
∴ `lambda = 36/(5 xx 1.097 xx 10^7)`
= `(36 xx 10^-7)/5.485`
= 6.563 × 10−7 m
= 6563 Å
(b) For the second line in Balmer series, n = 2, m = 4
∴ `1/lambda_2 = R_H(1/(2)^2 - 1/(4)^2)`
= `R_H (1/4 - 1/16)`
∴ `1/lambda_2 = R_H3/16`
∴ `lambda_2 = 16/(3 R_H)`
= `16/(3 xx 1.097 xx 10^7)`
= `16/3.291 xx 10^-7`
= 4.862 × 10−7 m
= 4862 Å
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