Question

What is the wave number of 4th line in balmer series of hydrogen spectrum. (R =1,09,677 cm inverse)

Answer 1

wave number = R*Z²*[1/(n1)²-1/(n2)²]

here n1 =2 , n2 =6

wave number =1.1*10^(7)[1/(2)²-1/(6)²] = 2.44*10^(6)

Answer 2

The wave number of Balmer series of lines in the hydrogen emission spectrum is

wave number = $R_{H}$[$\frac{1}{2²} -\frac{1}{n_{i}² }$]

Where $n_{i}$=3,4,5,6.... and $R_{H}$ = Rydberg constant

In the this quation,

$n_{i}$ = 6

Hence,

wave number = 109677[$\frac{1}{2²}$ ₋ $\frac{1}{n_{6} ²} }$]

                                               = 109677[$\frac{1}{4}$ ₋$\frac{1}{36}$]

                                              = 109677×$\frac{2}{9}$

                 wave number      =    24128.94 cm⁻