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

Here, `lambda=970Å`
`E=(hc)/lambda=12375/970 eV=12.7eV`
After excitation, let electron jump to nth state
`:. -13.6/(n^(2))=-13.6+12.7=0.9`
`n^(2)=13.6/0.9=15.1~=16`
`:. n=4`
Total number of lines in the emission spectrum
`=(n(n-1))/2=(4(4-1))/2=6`