The ground state energy of hydrogen atom is -13.6 eV.
(i) What is the kinetic energy of an electron in the `2^(nd)` excited state ?
(ii) If the electron jumps to the ground state from the `2^(nd)` excited state, calculate the wavelength of the spectral line emitted.

Explanation

The energy of an electron in nth orbit is given by
`E_(n)=(-13.6)/(n^(2))eV`
(ii) For `2 nd` excited state , `n=3`
`:. E_(3)=(-13.6)/((3)^(2))=(-13.6)/(9)=-1.51 eV`
Required energyf to jump electron to the ground state from the `2 nd` excited state
`E=E_(3)-E_(1)`
`=(-13.6)/((3)^(2))= -((-13.6)/((1)^(2)))= -1.51+13.6=12.09 eV`
`:.` Wavelength of the spectral line emitted is
`lambda=(hc)/(E )[ because E=(hc)/(lambda)]`
`lambda=(6.626xx10^(-34)xx3xx10^(8))/(12.09xx1.6xx10^(-19))=(19.878xx10^(-7))/(19.344)=1.027xx10^(-7)m= 1027Å`