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.

Explanation

Given, Bohr.s radius in ground state , (n=1), `r_0`=0.53 Å =0.53 x `10^(-10)` m
Bohr.s radius in excited state (n),`r_n` =2.12Å
`=2.12xx10^(-10)` m
We know that
`r_n prop n^2`
`therefore (n_2/n_1)^2=r_2/r_1`
`(n/1)^2=r_n/r_0=(2.12xx10^(-11))/(0.53xx10^(-11))=4`
`rArr n^2`=4 or n=2
Also, energy in this state is given by
`E_n=(-13.6)/n^2` eV=`(-13.6)/2^2` =-3.4 eV