Question

At what energy level does the Bohr hydrogen atom have diameter 8.57 nm ?

Answer 1

Radius of the atom is r = n² * h² * Є₀/π * m * e² ….( 1 )

it is given that diameter D = 8.57 nm = 8.57 * 10^-9 m

So radius is r = D/2 = 8.57 * 10^-9 m/2 = 4.285 * 10^-9 m

Here we can find the energy level simply by substituting value of r in the equation ( 1 ) which gives n = 9

But we can also calculate it by calculating the energy of the energy level

Energy of hydrogen atom in an energy level n is given by

                    E = - m * e⁴/8 * n² * h² * Є₀² ………. ( 2)       

From ( 1 ) and ( 2 ) we can observe that

                   E = ( - e²/8 π * Є₀ ) * ( π * m * e²/n² * h² * Є₀ )

                      = ( - e²/8 π Є₀ * r) ……… ( 3 )

                E = - ( 1.6 * 10^-19 )² / ( 8 * 3.14 * 8.85 * 10^-12 * 4.285 * 10^-9 )

                   = - ( 2.56 * 10^-38/952.60 * 10^-21 )

                   = - 0.00269 * 10^-17 J

                   = - ( 0.00269 * 10^-17/ 1.6 * 10^-19 ) eV

                  = - 0.168 eV

Now energy of an hydrogen is given by E = - 13.6 eV/n²

             So           - 0.168 = - 13.6 /n²

               Then            n² = 13.6/0.168 = 80.9 = 81 ( appr )

                Finally        n = 9

So the hydrogen atom is in the energy level n = 9