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.)
The energy of an emitted photon for an electronic transition in the hydrogen atom is given by the equation
E = E0*(1/ni2 – 1/nf2)
where ni and nf 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
E1→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).
Calculate the energy difference (E) for the transition from n = 1 to n = 6...
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