The ground state energy of hydrogen atom is -13.6 eV . If an electron makes a transition from an energy level -1.51 eV to -3.4 eV, calculate the wavelength of the spectrum line emitted and name the series of hydrogen spectrum to which it belongs.

Explanation

The energy in the n-th level of hydrogen atom,
`E_n=-(13.6)/(n^2)`
`"For energy" -1.51 eV,`
`-1.51 =-(13.6)/(n^2) or, n^2 =(13.6)/(1.51)~~9 or, n=3`
`"For energy" -3.4 eV`,
`-3.4 =-(13.6)/(n^2) or, n^2 =-(13.6)/(-3.4)~~4 or, n=2`
Thus, an electron makes transition from energy level n=3 to n=2.
`therefore (hc)/(lambda_(32))=(me^4)/(8 in_0^2h^2)(1/(n_2^2)-1/(n_3^2))`
`or, (hc)/(lambda_(32))=21.76xx10^(-19)(1/(2^2)-1/(3^2))`
`or, lambda_(32)=(hc)/(21.76xx10^(-19)(1/4-1/9))`
`=(6.62xx10^(-34)xx3xx10^(8))/(21.76xx10^(-19))xx(36)/5=6.57xx10^(-7)m`
This wavelength belongs to Balmer series.