Derive an expression for the frequency of spectral series by assuming the expression for the total energy of the election of hydrogen.

Explanation

When an atom in the excited state returns to the normal state, the transition results in the radiation of energy .
`E_(2)-E_(1) =hv=(hc)/(lamda)`
i.e., `(1)/(lamda)=((1)/(hc))(E_(2)-E_(1))`
where, `E_(n) = (-Z^(2)me^(4))/(8epsilon_(0)^(2)n^(2)h^(2))`
hence `1/lamda = (-Z^(2)me^(4))/(8epsilon_(0)^(2)h^(3)c)(-1/(n_(2)^(2))( -1/n_(1)^(2)))`
i.e.
`1/lamda = (-Z^(2)me^(4))/(8epsilon_(0)^(2)h^(3)c)(-1/(n_(1)^(2))- 1/(n_(2)^(2)))`
i.e., `1/lamda=Z^(2)R(1/n_(1)^(2)-1/(n_(2)^(2)))`
where, `R=(me^(4))/(8epsilon_(0)^(2)h^(3)c)` is called Rydberg's constant.
for `H_(2)` atom , Z=1
`:.` Wave number of hydrogen atom, `bar(v)=1/lamda=R((1)/(n_(1)^(2))-1/n_(2)^(2))`
Multiplying of radiation `v=Rc((1)/(n_(1)^(2))-1/n_(2)^(2)) " where " n_(2)=n_(1)+1,n_(1)+2,n_(1)+3,...oo`