Question

An Electron initially in the n=2 state of a hydrogen atom is excited by a photon to the n=5 state.

a. What is the energy of the excitatoin photon?

b. As the atom relaxes, the electron transitions to the ground state. What is the energy of the photon released during the electron transition?

c. What is the frequency of the released photon?

d. What is the name of the scientist who first modeled the atom as a miniature solar system?

Answer 1

Energy of the eelctron in n=2 state of the hydrogen is E = -13.6 eV / 2 ²

                                                                                 = -3.4 eV

Energy of the electron in n=5 state of the hydrogen is E ' = -13.6 eV / 5 ²

                                                                                  = -0.544 eV

(a).The energy of the excitatoin photon E " = E ' - E

                                                              = 2.856 eV

(b). The electron transitions to the ground state then the energy of the photon released during the electron transition is E 1 = ?

We know the energy of the electron in the ground state i.e., n = 1 state is E " = -13.6 eV / 1 ²

                      E " = -13.6 eV

Therefore E ₁ = E ' - E "

                    = -0.544 eV - (-13.6 eV )

                   = 13.056 eV

(c). The frequency of the released photon f = E ₁ / h

Where h = planck's constant = 6.625 x 10 -34 J s

Substitute values you get   f = [13.056 x 1.6 x 10 ^-19 J ] / [6.625 x 10 ^-34 J s ]

                                         = 3.15 x 10 ¹⁵ Hz

(d).The name of the scientist who first modeled the atom as a miniature solar system Rutherford