The Workbook

1. The electron moves around the nucleus only in certain allowed circular orbits. In these orbits it does not radiate energy, even though it accelerates.

2. The allowed orbits are those in which the angular momentum is a whole multiple of \(\dfrac{h}{2\pi}\): \(mvr = \dfrac{nh}{2\pi}\).

3. The atom radiates only when the electron jumps from a higher orbit to a lower one. The photon carries the energy difference: \(h\nu = E_i - E_f\).

\(L = \dfrac{nh}{2\pi}\). For \(n = 1\): \(L = 1.05 \times 10^{-34}\) J s. For \(n = 3\): \(L = 3 \times 1.05 \times 10^{-34} = 3.15 \times 10^{-34}\) J s.

\(r_n = 0.53\dfrac{n^2}{Z}\) Å.

1. \(Z = 1\): \(r_1 = 0.53\) Å, \(r_2 = 0.53 \times 4 = 2.12\) Å, \(r_3 = 0.53 \times 9 = 4.77\) Å.

2. \(Z = 2\), \(n = 2\): \(r = 0.53 \times \dfrac{4}{2} = 1.06\) Å.

3. \(Z = 3\), \(n = 3\): \(r = 0.53 \times \dfrac{9}{3} = 1.59\) Å.

1. \(v = 2.2 \times 10^{6} \times \dfrac{1}{2} = 1.1 \times 10^{6}\) m/s.

2. \(v = 2.2 \times 10^{6} \times \dfrac{2}{1} = 4.4 \times 10^{6}\) m/s.

Ground state of hydrogen: \(\dfrac{v_1}{c} = \dfrac{2.2 \times 10^{6}}{3 \times 10^{8}} = 0.0073 \approx \dfrac{1}{137}\). The speed is far below \(c\), so Newton's mechanics is good enough here.

1. From \(mvr = \dfrac{nh}{2\pi}\): \(r = \dfrac{nh}{2\pi mv}\). Put this in \(mv^2 = \dfrac{kZe^2}{r}\):

   \(mv^2 = kZe^2 \times \dfrac{2\pi mv}{nh}\), so \(v = \dfrac{2\pi kZe^2}{nh}\). Everything except \(Z\) and \(n\) is constant: \(v \propto \dfrac{Z}{n}\).

2. \(r = \dfrac{nh}{2\pi mv} = \dfrac{nh}{2\pi m} \times \dfrac{nh}{2\pi kZe^2} = \dfrac{n^2h^2}{4\pi^2 mkZe^2}\). So \(r \propto \dfrac{n^2}{Z}\).

1. \(T_1 = \dfrac{2\pi r_1}{v_1} = \dfrac{2\pi \times 0.53 \times 10^{-10}}{2.2 \times 10^{6}} = 1.5 \times 10^{-16}\) s.

   Frequency \(= \dfrac{1}{T_1} \approx 6.6 \times 10^{15}\) revolutions per second.

2. \(T = \dfrac{2\pi r}{v} \propto \dfrac{n^2/Z}{Z/n} = \dfrac{n^3}{Z^2}\).

   For \(n = 2\): \(T_2 = 2^3 T_1 = 8 \times 1.5 \times 10^{-16} = 1.2 \times 10^{-15}\) s.

1. \(E_n = -\dfrac{13.6}{n^2}\) eV: \(E_1 = -13.6\) eV, \(E_2 = -3.4\) eV, \(E_3 = -1.51\) eV, \(E_4 = -0.85\) eV.

2. In a Bohr orbit \(K = -E\) and \(U = 2E\). For \(n = 2\): \(K = 3.4\) eV and \(U = -6.8\) eV. Check: \(K + U = -3.4\) eV \(= E_2\).

Ionization energy \(= 0 - E_1 = 13.6Z^2\) eV.

H: \(13.6\) eV. He\(^+\): \(13.6 \times 4 = 54.4\) eV. Li\(^{2+}\): \(13.6 \times 9 = 122.4\) eV.

The ionization potential is the potential difference that gives an electron this energy: \(13.6\) V for hydrogen.

1. \(E_2 - E_1 = -3.4 - (-13.6) = 10.2\) eV.

2. \(E_3 - E_1 = -1.51 + 13.6 = 12.09\) eV.

3. For He\(^+\) every level is \(Z^2 = 4\) times deeper: \(4 \times 10.2 = 40.8\) eV.

1. A (\(3 \to 2\)): \(-1.51 - (-3.4) = 1.89\) eV.

   B (\(4 \to 1\)): \(-0.85 + 13.6 = 12.75\) eV.

   C (\(2 \to 1\)): \(-3.4 + 13.6 = 10.2\) eV.

2. \(\lambda = \dfrac{1240}{E}\) nm: A: \(656\) nm. B: \(97\) nm. C: \(122\) nm.

3. Only A (\(656\) nm, red) is visible. A ends on \(n = 2\): Balmer series. B and C end on \(n = 1\): Lyman series, in the ultraviolet.

\(E = E_3 - E_1 = -1.51 + 13.6 = 12.09\) eV. \(\lambda = \dfrac{1240}{12.09} \approx 103\) nm (ultraviolet).

Balmer \(\alpha\): \(E = 1.89\) eV, \(\lambda = \dfrac{1240}{1.89} \approx 656\) nm (red).

Lyman \(\alpha\): \(E = 10.2\) eV, \(\lambda = \dfrac{1240}{10.2} \approx 122\) nm (ultraviolet).

1. \(H_\alpha\) is at \(656\) nm and \(H_\beta\) at \(486\) nm.

2. All Balmer lines end on \(n = 2\). \(H_\alpha\): \(3 \to 2\). \(H_\beta\): \(4 \to 2\). \(H_\gamma\): \(5 \to 2\). \(H_\delta\): \(6 \to 2\).

3. The limit is \(\infty \to 2\), an electron falling from the edge of the atom. The upper levels \(E_n = -13.6/n^2\) get closer and closer together as \(n\) grows, so the photon energies, and the lines, crowd together.

The shortest wavelength has \(n_2 = \infty\): \(\dfrac{1}{\lambda} = \dfrac{R}{n_1^2}\), so \(\lambda = \dfrac{n_1^2}{R}\) with \(\dfrac{1}{R} = 91.2\) nm.

1. Lyman, \(n_1 = 1\): \(91.2\) nm.

2. Balmer, \(n_1 = 2\): \(4 \times 91.2 = 365\) nm.

3. Paschen, \(n_1 = 3\): \(9 \times 91.2 = 820\) nm.

The longest wavelength is the smallest jump, \(4 \to 3\): \(\dfrac{1}{\lambda} = R\left(\dfrac{1}{9} - \dfrac{1}{16}\right) = \dfrac{7R}{144}\).

\(\lambda = \dfrac{144}{7R} = \dfrac{144}{7} \times 91.2 \approx 1875\) nm. This is in the infrared.

Lyman \(\alpha\) (\(2 \to 1\)): \(\dfrac{1}{\lambda_L} = R\left(1 - \dfrac{1}{4}\right) = \dfrac{3R}{4}\), so \(\lambda_L = \dfrac{4}{3R}\).

Balmer \(\alpha\) (\(3 \to 2\)): \(\dfrac{1}{\lambda_B} = R\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = \dfrac{5R}{36}\), so \(\lambda_B = \dfrac{36}{5R}\).

\(\dfrac{\lambda_L}{\lambda_B} = \dfrac{4}{3R} \times \dfrac{5R}{36} = \dfrac{20}{108} = \dfrac{5}{27}\).

For a hydrogen-like ion \(\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\), so \(\lambda \propto \dfrac{1}{Z^2}\) for the same transition.

He\(^+\) has \(Z = 2\): \(\lambda = \dfrac{656}{4} = 164\) nm.

1. \(n = 4\): \(\dfrac{4 \times 3}{2} = 6\) lines. \(n = 5\): \(\dfrac{5 \times 4}{2} = 10\) lines.

2. Balmer lines end on \(n = 2\). From \(n = 4\) these are \(4 \to 2\) and \(3 \to 2\): two lines. The other four are \(4 \to 3\) (Paschen) and \(4 \to 1\), \(3 \to 1\), \(2 \to 1\) (Lyman).

\(1.89\) eV is \(E_3 - E_2\), so the atom drops to \(n = 2\).

The second photon takes it to the ground state: \(E_2 - E_1 = 10.2\) eV. Check: \(1.89 + 10.2 = 12.09\) eV \(= E_3 - E_1\).

1. \(r_2 = 0.53 \times 4 = 2.12\) Å, \(r_3 = 0.53 \times 9 = 4.77\) Å. \(r_2 : r_3 = 4 : 9\), as drawn.

2. \(v \propto 1/n\), so \(v_3 : v_2 = 2 : 3\). The outer electron moves more slowly.

3. \(E = E_3 - E_2 = 1.89\) eV. \(\lambda = \dfrac{1240}{1.89} \approx 656\) nm, the red \(H_\alpha\) line.

1. \(H_\alpha\): \(\dfrac{1}{4} - \dfrac{1}{9} = 0.139\), the first point. Lyman \(\alpha\): \(1 - \dfrac{1}{4} = 0.75\), the dotted point.

2. Slope \(= \dfrac{8.23 \times 10^{6}}{0.75} \approx 1.1 \times 10^{7}\) m\(^{-1}\). This is the Rydberg constant \(R\); the graph is the Rydberg formula drawn as a straight line.

3. Bracket \(= 1\) means \(n_1 = 1\), \(n_2 = \infty\): \(\dfrac{1}{\lambda} = R\), \(\lambda = 91\) nm. This is the Lyman series limit.

1. \(122\), \(103\) and \(97\) nm are Lyman lines. The atom absorbs upward from the ground state: \(1 \to 2\) (\(10.2\) eV), \(1 \to 3\) (\(12.09\) eV) and \(1 \to 4\) (\(12.75\) eV).

2. A Balmer line is absorbed only by an atom already in \(n = 2\). In cold gas almost every atom is in the ground state \(n = 1\), so only jumps starting from \(n = 1\) (Lyman) are absorbed.

\(15\) eV is more than the ionization energy \(13.6\) eV. The electron is freed from the atom.

Its kinetic energy is the surplus: \(K = 15 - 13.6 = 1.4\) eV.

1. Put \(\lambda = \dfrac{h}{mv}\) in \(2\pi r = n\lambda\): \(2\pi r = \dfrac{nh}{mv}\), so \(mvr = \dfrac{nh}{2\pi}\). Bohr's rule says the orbit holds a whole number of de Broglie waves.

2. \(\lambda = 2\pi r_1 = 2\pi \times 0.53 = 3.33\) Å.

   Check: \(K = 13.6\) eV, \(\lambda = \dfrac{12.27}{\sqrt{13.6}} = \dfrac{12.27}{3.69} = 3.33\) Å. The two agree.

1. \(\mu = \dfrac{m \cdot m}{m + m} = \dfrac{m}{2}\).

2. \(\mu = \dfrac{m}{1 + m/M}\). When \(M \gg m\), \(m/M \to 0\) and \(\mu \to m\). In hydrogen the proton is \(1836\) times heavier. Treating the proton as fixed and using \(m\) is correct to about one part in \(1836\).

The centre of mass stays at rest, so \(m\,r_e = M\,r_p\). Then \(\dfrac{r_e}{r_p} = \dfrac{M}{m} = 1836\). The proton moves in a circle \(1836\) times smaller than the electron's.

1. Both masses are \(m\): \(\mu = \dfrac{m}{2}\).

2. \(E_n \propto \mu\), so every energy is halved: \(E_1 = -6.8\) eV. The ionization energy is \(6.8\) eV.

3. \(r_n \propto \dfrac{1}{\mu}\), so every radius is doubled: \(r_1 = 1.06\) Å.

Every level is halved, so every photon energy is halved and every wavelength is doubled.

\(2 \to 1\): \(E = \dfrac{10.2}{2} = 5.1\) eV, \(\lambda = \dfrac{1240}{5.1} \approx 243\) nm (hydrogen: \(122\) nm).

\(3 \to 2\): \(E = \dfrac{1.89}{2} = 0.945\) eV, \(\lambda = \dfrac{1240}{0.945} \approx 1312\) nm (hydrogen: \(656\) nm).

1. \(\mu = \dfrac{207 \times 1836}{207 + 1836}\,m = \dfrac{380052}{2043}\,m \approx 186\,m\).

2. \(E_1 = -13.6 \times 186 \approx -2530\) eV \(= -2.53\) keV. The muon is bound about \(186\) times more tightly.

3. \(r_1 = \dfrac{0.53}{186} \approx 2.85 \times 10^{-3}\) Å \(= 2.85 \times 10^{-13}\) m. The muon orbits about \(200\) times closer to the proton than the electron does.

1. \(R_H = R_\infty \times \dfrac{1836}{1837} = 1.09737 \times 10^{7} \times 0.999456 = 1.09677 \times 10^{7}\) m\(^{-1}\).

2. Fractional change \(= \dfrac{1}{1837} = 5.4 \times 10^{-4}\), that is \(0.054\%\). Small, but spectrometers measure wavelengths far more precisely than this.

1. Deuterium's nucleus is heavier, so its reduced mass is a little larger. The level energies, and the photon energies, are larger in proportion to \(\mu\). A larger photon energy means a shorter wavelength.

2. \(\lambda \propto \dfrac{1}{E} \propto \dfrac{1}{\mu}\), so \(\dfrac{\lambda_H}{\lambda_D} = \dfrac{\mu_D}{\mu_H}\).

   \(\mu_H = \dfrac{1836}{1837}m\) and \(\mu_D = \dfrac{3672}{3673}m\). The ratio is \(\dfrac{3672 \times 1837}{3673 \times 1836} = 1.000272\).

3. \(\lambda_H - \lambda_D = \lambda_H\left(1 - \dfrac{1}{1.000272}\right) \approx 656.3 \times 2.72 \times 10^{-4} \approx 0.18\) nm. The figure shows this split.

1. \(p = \dfrac{E}{c} = \dfrac{10.2 \times 1.6 \times 10^{-19}}{3 \times 10^{8}} = 5.44 \times 10^{-27}\) kg m/s.

2. The total momentum was zero before, so it is zero after. The atom has momentum \(5.44 \times 10^{-27}\) kg m/s in the direction opposite to the photon. This is conservation of momentum.

\(v = \dfrac{p}{M} = \dfrac{5.44 \times 10^{-27}}{1.67 \times 10^{-27}} \approx 3.3\) m/s.

\(K = \dfrac{1}{2}Mv^2 = \dfrac{1}{2} \times 1.67 \times 10^{-27} \times (3.26)^2 \approx 8.9 \times 10^{-27}\) J.

In eV: \(\dfrac{8.9 \times 10^{-27}}{1.6 \times 10^{-19}} = 5.5 \times 10^{-8}\) eV.

Fraction of the photon energy: \(\dfrac{5.5 \times 10^{-8}}{10.2} \approx 5 \times 10^{-9}\). The recoil takes a negligible share.

1. Momentum is conserved. The photon carries momentum to the right, so the atom must carry an equal momentum to the left.

2. Less. The energy \(\Delta E\) released by the atom is shared: \(\Delta E = E + K\), where \(K\) is the recoil kinetic energy. So \(E = \Delta E - K\), slightly less than \(\Delta E\).

1. \(K = \dfrac{p^2}{2M} = \dfrac{(E/c)^2}{2M} = \dfrac{E^2}{2Mc^2}\).

2. \(K = \dfrac{(10.2)^2}{2 \times 939 \times 10^{6}} = \dfrac{104}{1.88 \times 10^{9}} \approx 5.5 \times 10^{-8}\) eV, the same as Question 10.

3. \(E = \Delta E - K\). Since \(K\) is tiny, put \(E \approx \Delta E\) inside \(K\): \(E \approx \Delta E - \dfrac{\Delta E^2}{2Mc^2} = \Delta E\left(1 - \dfrac{\Delta E}{2Mc^2}\right)\).

1. To the right, the direction of the photon. The photon's momentum \(E/c\) is transferred to the atom.

2. Slightly more. The photon must supply \(\Delta E\) for the excitation and also the recoil kinetic energy: \(E = \Delta E + K\).

3. \(p = E/c \approx 5.44 \times 10^{-27}\) kg m/s as before, so \(v = p/M \approx 3.3\) m/s.

1. Momentum must be conserved. The particle brought momentum, so something must carry it after the collision. Moving bodies have kinetic energy, so not all of \(K\) can go into excitation.

2. Before: momentum \(p = mv\), \(K = \dfrac{p^2}{2m}\). After: the pair of mass \((m + M)\) has the same \(p\), so \(K' = \dfrac{p^2}{2(m + M)}\).

   \(\dfrac{K'}{K} = \dfrac{m}{m + M}\), so \(K' = \dfrac{m}{m + M}K\).

3. Energy available for excitation \(= K - K' = \dfrac{M}{m + M}K\). At threshold this equals \(\Delta E\): \(K_{\min} = \Delta E\,\dfrac{m + M}{M}\).

\(m = M\), so \(K_{\min} = \Delta E\,\dfrac{2M}{M} = 2\Delta E = 2 \times 10.2 = 20.4\) eV. Half the kinetic energy stays as motion of the pair.

Again \(m \approx M\), so \(K_{\min} = 2\Delta E\).

1. \(2 \times 10.2 = 20.4\) eV.

2. Ionization needs \(\Delta E = 13.6\) eV: \(K_{\min} = 27.2\) eV.

\(K_{\min} = \Delta E\,\dfrac{m + M}{M} = 10.2 \times \dfrac{1837}{1836} \approx 10.2\) eV.

The electron is so light that the atom hardly moves after the collision. Almost all the electron's kinetic energy is available, so the threshold is nearly \(\Delta E\) itself.

1. With \(m = M\), the available energy is \(\dfrac{M}{m + M}K = \dfrac{25}{2} = 12.5\) eV, the bar in the figure.

2. \(12.5\) eV is above \(10.2\) eV (\(n = 2\)) and \(12.09\) eV (\(n = 3\)) but below \(12.75\) eV (\(n = 4\)). The atom can reach \(n = 2\) or \(n = 3\).

3. From \(n = 3\): \(3 \to 2\) (\(1.89\) eV, \(656\) nm), \(3 \to 1\) (\(12.09\) eV, \(103\) nm) and then \(2 \to 1\) (\(10.2\) eV, \(122\) nm). Three lines in all.

The total momentum is zero, so the atoms can stop dead and all their kinetic energy can go into excitation.

\(2K = 10.2\) eV, so \(K = 5.1\) eV each. In Question 15 the pair had to keep moving to conserve momentum, which wasted half the energy.

A photon is absorbed whole or not at all, and it disappears. Its whole energy must fit the gap \(\Delta E\) exactly (apart from the tiny recoil correction).

A particle survives the collision. It can give the atom exactly \(\Delta E\) and carry away the rest as its own kinetic energy. So any energy above the threshold will do.

1. \(m = 4\), \(M = 1\): \(K_{\min} = 10.2 \times \dfrac{4 + 1}{1} = 51\) eV. A heavy projectile on a light target wastes most of its energy in motion.

2. \(m = 1\), \(M = 4\): \(K_{\min} = 40.8 \times \dfrac{1 + 4}{4} = 51\) eV.

1. \(20.4\) eV is exactly the threshold, so the atoms move together after the collision. Kinetic energy left \(= \dfrac{m}{m + M}K = \dfrac{20.4}{2} = 10.2\) eV.

   The two atoms have the same mass and the same speed, so each has \(5.1\) eV.

2. The excited atom returns to \(n = 1\) and emits a \(10.2\) eV photon (Lyman \(\alpha\)).