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Questions
Calculate the shortest wavelength of electromagnetic radiation present in Balmer series of hydrogen spectrum.
Calculate the shortest wavelength of Balmer series.
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Solution
For shortest wavelength in Balmer series, nf = 2, ni = ∞
`1/lambda = R [1/n_f^2 - 1/n_i^2]`
`1/lambda = R(1/(2)^2 -0)`
`1/lambda = R/4`
∴ `lambda = 4/"R"`
∴ R = 1.097 × 107 m−1
` = 4/(1.097 xx 10^7) m`
` = 3.646 xx 10^-7 m`
= 3646 Å
= 364 nm
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