Assuming Bohr's atomic model find (a)largest wavelength in Balmer series, (b)the excitation energy of n=3 level of `He^+` atom.

Explanation

(a)In Balmer series electron makes transition from higher energy state to n=2 energy state. The wavelengths are given by
`1/lambda=R[1/2^2-1/n^2]`, where n=3,4,5….
To calculate the largest wavelength we need minimum energy difference . Hence, for largest wavelength we can put n=3.
`1/lambda=R[1/4-1/9]=(5R)/36`
`rArr lambda=36/(5R)=36/(5xx1.097xx10^7)=6.563xx10^(-7)` m =656.3 nm
(b)According to Bohr.s model we can write the energy of hydrogen like atoms as follows :
`E=-Z^2(13.6/n^2)`eV
Excitation energy for energy level n=3 is equal to change in energy when atom is shifted from n=1 to n=3 level. Hence energy required can be written as follows :
`E_3-E_1=(-2)^2xx13.6xx[1/3^2-1/1^2]`=48.35 eV