Calculate (a) the wavelength and (b) the frequencey of the `Hbeta`
line of the Balmer series for hydrogen.

Answer

A, D

Explanation

`Hbeta` line of Balmer series corresponds to the transition from `n = 4 to n = 2 level`
Using Eq. )xiii), the corresponding walelength for `Hbeta` line is `
`(1)/(lambda)` = (1.097xx10^7) ((1)/(2^2)-(1)/(4^2)) = 0.25056xx10^7`
`:. Lambda =4.9xx10^7m`
(b) `f = (c)/(lambda) = (3.0xx10^8)/(4.9xx10^(-7) = 6.12xx10^14 Hz`