Question

The energy of the n = 3 Bohr orbit is -13.6 eV for an unidentified ionized atom in which only one electron moves about the nucleus. What is the radius of the n = 6 orbit for this species?

Answer 1

given n = 3,

E = -13.6 eV

let z is the atomic number.

we know,

E = -13.6*z^2/n^2 eV

-13.6 eV = 13.6*z^2/n^2 eV

z^2 = n^2

z = n

= 3

radius of orbit, rn = 0.0529*n^2/z nm

r6 = 0.0529*6^2/3

= 6.348*10^-10 m or 0.6348 nm <<<<----Answer