Calculate the amount of energy released for transition of electrons associated with 1 gram atom of hydrogen so that lowest frequency spectral line is formed in the visible region of the spectrum. `[R_(H)=1.1xx10^(7)m^(-1)]`.

Explanation

The spectral line under consideration belongs to Balmer series. Now for the lowest frequency spectral line in Balmer series we have `n_(1)=2 and n_(2)=3`.
`therefore (1)/(lamda)=R_(H)[(1)/(2^(2))-(1)/(3^(2))]=1.1xx10^(7)x(5)/(36)`
Energy released in the transition of each electron,
`E=hv=hxx(c)/(lamda)`
`=6.626xx10^(-34)xx3xx10^(8)xx1.1xx10^(7)xx(5)/(36)`
`=3.037xx10^(-19)J`
So emission of energy associated with 1 gram atom of hydrogen`=N_(0)xxhv=(6.022xx10^(23))xx(3.037xx10^(-19))` ltbRgt =182.8kJ.