Problem#1
(a) Show that, as n gets very large, the energy levels of the hydrogen atom get closer and closer together in energy. (b) Do the radii of these energy levels also get closer together?
Answer:
En = –13.6 eV/n2
(a) En = –13.6 eV/n2 and En+1 = –13.6 eV/(n+1)2
∆E = En+1 – En = –13.6 eV{1/(n+1)2 – 1/n2}
∆E = –13.6 eV{(n2 – (n+1)2/(n2(n+1)2)}
∆E = (13.6eV)(2n+1)/[n2(n+1)2]
As n becomes large, ∆E → (13.6 eV)(2n/n4) = (13.6 eV)(2/n3)
Thus ∆E becomes small as n becomes large.
(b) rn = n2r1 so the orbits get farther apart in space as n increases.
There are an infinite number of bound levels for the hydrogen atom. As n increases the
energies of the bound levels converge to the ionization threshold.
Problem#2
(a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels. (b) Calculate the orbital period in each of these levels. (c) The average lifetime of the first excited level of a hydrogen atom is 1.0 x 10-8 s. In the Bohr model, how many orbits does an electron in the n = 2 level complete before returning to the ground level?
Answer:
The orbital period for state n is the circumference of the orbit divided by the orbital speed.
(a) vn = (1/ε0)(e2/2nh)
n = 1 → v1 = (1.60 x 10-19 C)2/[2ε0(6.63 x 10-34 J.s)] = 2.18 x 106 m/s
n = 2 → v2 = ½ v1 = 1.09 x 106 m/s.
n = 3 → v3 = v1/3= 7.27 x 105 m/s.
(b) orbital perod = 2πrn/vn = (2ε0n2h2/me2)/[(1/ε0)(e2/2nh)] = 4ε02n3h3/me4
n = 1 → T1 = 4ε02(6.63 x 10-34 J.s)3/[(9.11 x 10-31 kg)(1.60 x 10-19C)4] = 1.53 x 10-16 s
n = 2 → T2 = T1(2)3 = 1.22 x 10-15 s
n = 3 → T3 = T1(3)3 = 4.13 x 10-15 s
(c) number of orbits = (1.0 x 10-8 s)/(1.22 x 10-15 s) = 8.2 x 106
The orbital speed is proportional to 1/n the orbital radius is proportional to n2, and the
orbital period is proportional to n3.
Problem#3
The energy-level scheme for the hypothetical oneelectron element Searsium is shown in Fig. 1. The potential energy is taken to be zero for an electron at an infinite distance from the nucleus. (a)
How much energy (in electron volts) does it take to ionize an electron from the ground level? (b) An 18-eV photon is absorbed by a Searsium atom in its ground level. As the atom returns to its ground level, what possible energies can the emitted photons have? Assume that there can be transitions between all pairs of levels. (c) What will happen if a photon with an energy of 8 eV strikes a Searsium atom in its ground level? Why? (d) Photons emitted in the Searsium transitions n = 3 → n = 2 and n = 3 → n = 1 will eject photoelectrons from an unknown metal, but the photon emitted from the transition n = 4 → n = 3 will not. What are the limits (maximum and minimum possible values) of the work function of the metal?
Answer:
The ionization threshold is at E = 0. The energy of an absorbed photon equals the energy gained by the atom and the energy of an emitted photon equals the energy lost by the atom.
(a) ∆E = 0 – (–20 eV) = +20 eV
(b) When the atom in the n = 1 level absorbs an 18-eV photon, the final level of the atom is n = 4.
The possible transitions from n = 4 and corresponding photon energies are n = 4 → n = 3, 3eV;
n = 4 → n = 2, 8 eV; n = 4 → = n = 1, 1.18 eV. Once the atom has gone to the n = 3 level, the following
transitions can occur: n = 3 → n = 2, 5 eV; n = 3 → n = 1, 15 eV. Once the atom has gone to the n = 2
level, the following transition can occur: n = 2 → n = 1, 10 eV. The possible energies of emitted photons
are: 3 eV, 5 eV, 8 eV, 10 eV, 15 eV and 18 eV.
(c) There is no energy level 8 eV higher in energy than the ground state, so the photon cannot be absorbed.
(d) The photon energies for n = 3 → n = 2 and for n = 3 → n = 1 are 5 eV and 15 eV. The photon energy
for n = 4 → n = 3 is 3 eV. The work function must have a value between 3 eV and 5 eV.
The atom has discrete energy levels, so the energies of emitted or absorbed photons have
only certain discrete energies.
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