Question

In the ground state of hydrogen atom, its Bohr radius is `5.3xx10^(-11)m`. The atom is excited such that the radius becomes `21.2xx10^(-11)m`. Find the value of principal quantum number and total energy of the atom in excited state.

Answer 1

Here, `r_1=5.3xx10^(-11)m , n_1=1`
`r_2=21.2xx10^(-11)m, n_2=? ,E=?`
As `rpropn^2`
`:. ((n_2)/(n_1))^2=(r_2)/(r_1)=(21.2xx10^(-11))/(5.3xx10^(-11))=4,`
`n_2=2n_1=2xx1=2`
`E_2=(E_1)/(n^2)=(-13.6)/(2^2)=-3.4eV`