The first member of the Balmer series of hydrogen atom has wavelength of 656.3 nm. Calculate the wavelength and frequency of the second member of the same series. Given, `c=3xx10^(8)m//s`.

Explanation

`1/lambda_1=R(1/n_1^2-1/n_2^2)`
For I member of Balmer series, `n_1=2` & `n_2=3, lambda_1` = 6563 Å
`1/lambda_1=R(1/4-1/9)=(5R)/36`
For II member of Balmer series, `n_1`=2 and `n_2`=4
`1/lambda_1=R(1/4-1/16)=(3R)/16`
Arriving at `lambda_2`= 4861Å
`v=c//lambda_2=6.1xx10^14` Hz
Given `lambda_0`=6563 Å=`6.536xx10^(-7)` m
`lambda_beta` =? `f_beta` = ?
`c=3xx10^(8) ms^(-1)`
`rArr 1/lambda_1 =R(1/n_1^2-1/n_2^2)`
For I member of Balmer series, `n_1=2,n_2=3`
`therefore 1/(6xx563xx10^(-7))=R(1/2^2-1/3^2)`
i.e., `10^7/6.563=R(5/36)`
Hence `R=36/5xx10^7/6.563`
i.e., `R=1.097xx10^7 m^(-1)`
For II member of Balmer series
`x_1=2, x_2=4`
`therefore 1/lambda_2=1.097xx10^7(1/2^2-1/4^2)`
i.e., `1/lambda_2=(1.097xx10^7xx12)/64`
Hence `lambda_2=64/(1.097xx12xx10^7)` m
i.e., `lambda_2=4.8617xx10^(-7)` m
`lambda_2`=4861.7Å
and frequency , `Y_2=c/lambda_2=(3xx10^8)/(4.8617xx10^(-7))`
i.e., `Y_2=0.6171xx10^15` Hz = `6.171xx10^14` Hz