(a) The radius of the innermost electron orbit of a hydrogen atom is `5.3 xx 10^(-11)` m. Calculate its radius in n =2 orbit.
(b) The total energy of an electron in the second excited state of the hydrogen atom is -1.51 eV. Find out its (i) kinetic energy and (ii) potential energy in this state.

Explanation

(a)`r_2/r_1=(n_2/n_1)^2`
`r_1=5.3xx10^(-11)` m
`n_2=2, n_1=1`
`therefore r_2=r_1(n_2/n_1)^2 =5.3xx10^(-11)xx4/1`
`=21.2xx10^(-11)` m
(b)Kinetic energy =-(Total energy ) =-(-1.51) =1.51 eV
Potential energy = 2E = 2 x (-1.51) =-3.02 eV