Question

Calculate the shortest wavelength of electromagnetic radiation present in Balmer series of hydrogen spectrum.

Calculate the shortest wavelength of Balmer series.

Answer 1

For shortest wavelength in Balmer series, n_f = 2, n_i = ∞

`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