A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV . Find the energy and wavelength of photon emitted.

Explanation

Since the binding energy is given as 0.85 eV, therefore
`E_i=-0.85` eV
Let n be the initial binding state of the electron, then
`E_n=-13.6 Z^2/n^2` or -0.85=-13.6`Z^2/n^2`
or n=4
Let the electron now jumps to the energy level n where excitation energy is 10.2 eV. Since the excitation energy `DeltaE` is defined with respect to ground state , therefore ,
`DetlaE=13.6 Z^2[1/n_1^2-1/n_2^2]eV "or" " " 10.2=13.6xx1^2[1/1^2-1/n^2]`
`n_f=2`
So, the electron makes a transition from energy level n=4 to n=2. Thus, the energy released is `DeltaE=E_4-E_2`
`DeltaE=13.6[1/2^2-1/4^2]`=2.55 V
`lambda="hC"/(DeltaE)=12400/"2.25eV"=5511Å`