Calculate the energy emitted when electron of 1.0 gm atom of Hydrogen undergo transition giving the spectrtal lines of lowest energy is visible region of its atomic spectra. Given that, `R_(H)`=`1.1xx10^(7) m^(-1)`,` c=3xx10^8m//sec`,` h=6.625xx10^(-34) Jsec`.

Explanation

Visible region of H-sepctrun correspond to Balmer series` n_(1)=2,n_(2)=3`(for minimium energy trasition )
`1//lambda=R_(H)=[(1)/(n_(1)^(2))-(1)/(n_(2)^(2)]]rArr(1)/lambda=R_(H)[(1)/(2^(2))-(1)/(3^(2)]]rArr(1)/lambda=1.1xx10^(7)[(1)/4-(1)/9]`
` [Rh =109677cm^(-1)=109677xx100m^(-1)approx1.1" "10^(7)m^(-1)] `
`lambda=6.55xx10^(-7)m`
`E=(hc)/lambda=(6.625xx10^(-34)xx3xx10^(8))/(6.55xx10^(-7))=3.03xx10^(-19)"Joule "`
` therefore`Energy released by 1gm atom of `H(1mol)`
` =3.03xx10^(-19)xx 6.023 xx10^(-23)` =`18.25xx 10^(4)J`=182.5 K