Using Bohr's formula for energy quantization, determine (i) the longest wavelength in Lymann series of hydrogen atom spectrum. (ii) the excitation energy of the n=3 level of `He^(+)` atom. (iii) the ionization potential of the ground state of `Li^(++)` atom.

Explanation

According to Bohr, the wavelength emitted when an electron jumps form `n_(1)th` to `n_(2)th` orbit is
`E=(hc)/lambda=E_2-E_1=(2pi^2mK^2e^4)/(h^2) [1/(n_1^2)-1/(n_2^2)]`
or `1/lambda=(2pi^2mK^2e^4)/(ch^3) (1/(n_1^2)-1/(n_2^2))`
`lambda=(ch^3)/(2pi^2mK^2e^(4) (1/(n_1^2)-1/(n_2^2)))`
(i) In Lyman series, `n_1=1`, for longest wave length, `n_2=2`
`:. lambda=(ch^3)/(2pi^2mK^2e^(4) (1-1/4))=(4ch^(3))/(6pi^2mK^2e^(4))`
`(4xx3xx10^8(6.6xx10^(-34))^3)/(6xx(22//7)^2xx9.1xx10^(-31)xx(9xx10^9)^2(1.6xx10^(-19))^4)`
`=1.218xx10^-7m=1218Å`
(ii) Energy required to excited the electron form the ground state `(n=1)` to `n=3` level of `(He^(+))` atom `(Z=2)` is
` E_3-E_1=(2pi^2mZ^2K^2e^4)/(h^2) (1/(1^2)-1/(3^2))`
`=(2xx(22//7)^2xx9.1xx10^(-31)xx4xx(9xx10^9)^2(1.6xx10^(-19))^4)/((6.6xx10^(-34))^2) xx8/9`
`=77.44xx10^(-19)"joule"`
`=(77.44xx10^(-19))/(1.6xx10^(-19))eV=48.4eV`
(iii) Ionisation energy of ground state of `Li^(++)` atom. Li atom has 3 electrons, out of which 2 are in first orbit and 1 is in 2nd orbit. `Li^(++)` atom is doubly ionised i.e., two electrons are already removed. The third electron has to be removed form `n_(1)=1` to `n_(2)=oo`.
`E=(2pi^2mZ^2K^2e^4)/(h^2) [1/(1^(2))-1/(oo^(2))]=122.7eV`