Hydrogen atom in ground state is excited by a monochromatic radiation of $\lambda =975 Å.$ Number of spectral lines in the resulting spectrum emitted will be:

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

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right)$
$\frac{1}{975}=R\left(\frac{1}{1}-\frac{1}{{\mathrm{n}}_2^2}\right)$
$\left(\because {\mathrm{n}}_1=1\right)$for ground state

where $\mathrm{R}$ is the Rydberg constant$=1.09\times {10}^7 {\mathrm{m}}^{-1}$
Solving above equation, we get rounded numerical value ${\mathrm{n}}_2=4$ (highest principal quantum number)
 

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

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