In the ground state of hydrogen atom , its Bohr radius is given as `5.3 xx 10^(-11)` m. The atom is excited such athat th radius becomes `21.2 xx 10^(-11)` m. Find (i) the value of principal quantum number , and (ii) the total energy of the atom in this excited state.

Explanation

Here `a_(0)` (or `r_(1)`) `=5.3 xx 10^(-11) m` and `r_(n) = 21.2 xx 10^(-11) m`
(i) `because r_(n) = n^(2) a_(0) rArr n = sqrt((r_(n))/(a_(0)) = sqrt((21.2xx10^(-11))/(5.3 xx 10^(-11))) = 2`
(ii) Total energy of atom in the excited state correspoding to n =2 .
`E_(n) = - (13.6)/(n^(2)) eV = - (13.6)/((2)^(2)) eV = -3.4 eV`