Hydrogen atom in ground state is excited by a monochromatic...

Given weavelength is $\lambda = 975 Å$
By Rydberg's equation for hydrogen atom is

$\dfrac{1}{\lambda} = R$\left(1{n}_{1}^{2} - 1{n}_{2}^{2}\right)
\dfrac{1}{975} =$R\left(\dfrac{1}{1} -1{n}_{2}^{2}\right)$
\left(∵ {n}_{1} = 1\right)$for ground state</p><p>where$R is the Rydberg constant$= 1 . 09 \times {10}^{7}{m}^{- 1}$
Solving above equation, we get rounded numerical value ${n}_{2} = 4$ (highest principal quantum number)

Total number of spectral lines is given in terms of highest principal quantum number as

$= \dfrac{n ( n - 1 )}{2} = \dfrac{4 ( 4 - 1 )}{2} =6$