A hydrogen atom in its ground state is irradiated by light of wavelength `970Å` Taking `hc//e = 1.237 xx 10^(-6)`eV m and the ground state energy of hydrogen atom as ` - 13.6 eV` the number of lines present in the emmission spectrum is

Answer

6

Explanation

Absorbed quantum energy in `(eV)=(hc)/(elamda)`
`=(1.237xx10^(-6))/(970xx10^(-10))`
`=12.75eV`
Energy of nth shell `=-13.6+12.75`
`=0.85`eV
We know
`E_(n)=(E_(1))/(n_(2))`
`thereforen^(2)=(E_(1))/(E_(n))=(-13.6)/(-0.85)=16`
i.e., `n=4`
Number of spectral emission lines `=(n(n-1))/(2)=(4xx3)/(2)=6`