1. Hydrogen: For an electron in the lowest energy it can orbit around proton, they have a separation of 5.3 *10-11 m. If you have a 4.5*10-19J photon (bit of light) hit the electron, it will transfer all of its energy to the proton electron interaction and the electron will start orbiting at a larger radius. Assuming all the energy went into the potential energy, what is the new distance between the electron and proton.

 rₙ = 1,325 10⁻⁹ m

Explanation:

To solve this problem we use the bohr atomic model

           Eₙ = -13.606 /n²     [eV]

the brackets indicate that the units are in electron volts.

let's reduce the photon energy to eV

        E = 4.5 10-19J (1 eV / 1.6 10⁻¹⁹ eV) = 2.8125 eV

This energy is in the visible range, so the transition must occur in this range, this is for the Balmer series whose initial number is n₀ = 2

for an atomic transition on two levels

         ΔE = Eₙ - E₀ =

         2.8125 =

          = 3.4015 - 2.8125 = 0.589

           n² = 13.606 / 0.589

           n² = 23.1

           n = 4.8

as n must be an integer

           n = 5

taking the quantum number as far as the electron goes, we substitute in the equation for the radius

           rn = n² a₀

where ao is the radius of the lowest level a₀ = 5.3 10⁻¹¹ m

            rₙ = 5 2 5.3 10⁻¹¹

            rₙ = 132.5 10⁻¹¹ m

            rₙ = 1,325 10⁻⁹ m

multiple

frequency or wavelenth of em radiation in a vacuum