The energy of the electron in the second and third Bohr's orbitals of the hydrogen atom is `-5.42 xx 10^(-12) erg` and `-2.42 xx 10^(-12) erg` respectively ,Calculate the wavelength of the emitted radiation when the electron drops from the third to the second orbit.

Explanation

`E_(3)` for `H = - 2.14 xx 10^(-12)` erg
`E_(2)` for `H = - 5.42 xx 10^(-12)` erg
For a jump from third to second shell
`Delta E = E_(3) - E_(2) = (hc)/(lambda)`
`:. Lambda = (hc)/(E_(3) - E_(2))`
`= (6.625 xx 10^(-27) xx 3 xx 10^(10))/( - 2.41 xx 10^(-8) cm = 6603Å`