Question

Calculate the energy difference (E) for the transition from n = 1 to n = 6 energy levels of hydrogen per 1 mol of H atoms. (Report your answer to at least 3 significant figures.)

Answer 1

The energy of an emitted photon for an electronic transition in the hydrogen atom is given by the equation

E = E₀*(1/n_i² – 1/n_f²)

where n_i and n_f are the principal quantum numbers of the initial and the excited states of the hydrogen atom and

E₀ = 13.6 eV is the energy of the photon in the ground state.

Plug in values and determine E₆ as below.

E₆ = (13.6 eV)*[1/(1)² – 1/(6)²]

= (13.6 eV)*(1 – 1/36)

= (13.6 eV)*(35/36)

= 13.2 eV

The energy difference of the transition is given as

E_1→6 = E₆ – E₀

= (13.2 eV) – (13.6 eV)

= -0.4 eV.

= (-0.4 eV)*(1.602*10^-19 J)/(1 eV)

= -6.408*10^-20 J

The above is the energy difference of a single electron in the transition of a hydrogen atom from n = 1 to n = 6 energy level.

Since 1 mole of hydrogen atoms are excited, hence, 1 mole of electrons are produced. Therefore, the energy difference of 1 mole of electrons is given as

ΔE = (6.02*10²³ electrons/mole)*(-6.408*10^-20 J/electron)

= -38576.16 J/mol

= (-38576.16 J)*(1 kJ)/(1000 J)

= -38.57616 kJ/mol

≈ -38.6 kJ/mol (ans).