The radio telescope at Ooty in the Nilgiri hills looks at clouds of thin gas between the stars. From these clouds it receives faint radio signals near \(327\,\text{MHz}\). They come from single atoms whose electrons jump between orbits with \(n\) close to 272. An atom in such a state is tens of thousands of times larger than an ordinary atom.
Two equations for one orbit
Bohr's model works for any atom with a single electron. Such an atom is a hydrogen-like atom: a nucleus of charge \(Ze\) and one electron. Hydrogen itself has \(Z = 1\). The helium ion \(\text{He}^+\) has \(Z = 2\), and the lithium ion \(\text{Li}^{2+}\) has \(Z = 3\). Atoms with two or more electrons are outside the model.
Two equations fix each orbit. First, the electric pull of the nucleus, \(kZe^2/r^2\), supplies the centripetal force \(mv^2/r\), as in the chapter on Dynamics of Circular Motion. Second, Bohr's rule fixes the angular momentum, \(mvr = nh/2\pi\). Two equations in two unknowns, \(v\) and \(r\), give both.
\[ mv^2 = \dfrac{kZe^2}{r} \]
Why. multiply the force equation \(mv^2/r = kZe^2/r^2\) by \(r\).
\[ v = \dfrac{kZe^2}{mvr} = \dfrac{2\pi kZe^2}{nh} \]
Why. divide step 1 by \(mvr\), then put \(mvr = nh/2\pi\) from Bohr's rule.
\[ r = \dfrac{nh}{2\pi mv} = \dfrac{n^2h^2}{4\pi^2 mkZe^2} \]
Read it. Bohr's rule gives \(r = nh/2\pi mv\); put in the speed from step 2.
Everything in these results is a constant except \(n\) and \(Z\). So the speed is proportional to \(Z/n\) and the radius to \(n^2/Z\). The radius of the lowest orbit of hydrogen is the Bohr radius, \(a_0 = 0.529\,\text{Å}\), where \(1\,\text{Å} = 10^{-10}\,\text{m}\). The speed on that orbit is \(2.18 \times 10^6\,\text{m/s}\).
Fig 3.1 shows the three trends together. The radii grow as \(1 : 4 : 9\), so the orbits spread apart quickly. The speeds fall as \(1 : \tfrac12 : \tfrac13\), because a weaker pull far out can hold only a slower circular motion. Even the fastest, \(v_1\), is only about \(c/137\), where \(c\) is the speed of light. So Newton's mechanics is good enough inside the hydrogen atom.
Find the radius of the orbit and the speed of the electron for (a) hydrogen, \(n = 2\); (b) \(\text{He}^+\), \(n = 2\); (c) \(\text{Li}^{2+}\), \(n = 3\). Show also that \(v_1/c\) is about \(1/137\).
Use \(r_n = 0.529\,n^2/Z\) Å and \(v_n = 2.18 \times 10^6\,Z/n\) m/s, with \(Z = 1\), \(2\) and \(3\).
(a) \(n^2/Z = 4\), \(Z/n = \tfrac12\). (b) \(n^2/Z = 2\), \(Z/n = 1\). (c) \(n^2/Z = 3\), \(Z/n = 1\).
(a) \(r = 2.12\,\text{Å}\), \(v = 1.09 \times 10^6\,\text{m/s}\). (b) \(r = 1.06\,\text{Å}\), \(v = 2.18 \times 10^6\,\text{m/s}\). (c) \(r = 1.59\,\text{Å}\), \(v = 2.18 \times 10^6\,\text{m/s}\).
\(v_1/c = 2.18 \times 10^6 / 3 \times 10^8 = 0.0073 \approx 1/137\). Values used: \(r_1 = 0.529\,\text{Å}\), \(v_1 = 2.18 \times 10^6\,\text{m/s}\), \(c = 3 \times 10^8\,\text{m/s}\).
The two numbers can be checked with the values of this unit. Write \(\hbar = h/2\pi\), so \(\hbar c = 1240/2\pi = 197.4\,\text{eV nm}\), and \(mc^2 = 0.512\,\text{MeV}\) for the electron. Then \(r_1 = (\hbar c)^2/(mc^2 \cdot ke^2) = 197.4^2/(512000 \times 1.44) = 0.0529\,\text{nm}\). And \(v_1/c = ke^2/\hbar c = 1.44/197.4 = 1/137\).
For a fixed nucleus, how does the speed of the electron depend on the quantum number \(n\)?
Find the radius of the lowest orbit, \(n = 1\), of the helium ion \(\text{He}^+\) (\(Z = 2\)), in ångström. Take \(r_1 = 0.529\,\text{Å}\) for hydrogen.
The energy of an orbit
The kinetic energy follows from the force equation, \(mv^2 = kZe^2/r\). So \(K = \tfrac12 mv^2 = kZe^2/2r\). The potential energy of the electron and the nucleus is \(U = -kZe^2/r\), negative because the charges attract, with \(U = 0\) when they are far apart. The total energy is the sum.
\[ E = K + U = \dfrac{kZe^2}{2r} - \dfrac{kZe^2}{r} = -\dfrac{kZe^2}{2r} \]
So on every orbit, \(K = -E\) and \(U = 2E\). You met the same three relations for a satellite in the chapter on Gravitational Energy and Orbits, since both forces fall as \(1/r^2\).
\[ E = -\dfrac{kZe^2}{2r_n} \]
Why. the total energy on a circular orbit, found above.
\[ E_1 = -\dfrac{ke^2}{2a_0} = -\dfrac{1.44\,\text{eV nm}}{2 \times 0.0529\,\text{nm}} = -13.6\,\text{eV} \]
Why. for hydrogen's lowest orbit, \(Z = 1\) and \(r = a_0 = 0.0529\,\text{nm}\).
\[ E_n = -\dfrac{kZe^2}{2a_0}\cdot\dfrac{Z}{n^2} = -13.6\,\dfrac{Z^2}{n^2}\ \text{eV} \]
Read it. put \(r_n = a_0 n^2/Z\) in step 1. One \(Z\) comes from the charge and one from the smaller radius.
The electron volt used here is the energy an electron gains in crossing a potential difference of one volt, \(1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}\). The chapter on Photons and the Photoelectric Effect uses it throughout. Atomic energies are a few electron volts, so it is the natural unit here.
The total energy is negative because the electron is bound. The zero of energy is an electron at rest far from the nucleus. A bound electron sits below that zero. To pull it out and leave it at rest far away, you must supply \(|E_n|\). That makes \(|E_n|\) the binding energy of the electron on orbit \(n\).
Find the total, kinetic and potential energies of the electron of hydrogen on the orbit \(n = 2\).
\(Z = 1\), \(n = 2\). The total energy comes from \(E_n = -13.6\,Z^2/n^2\) eV.
\(E_2 = -13.6/4\), then \(K = -E_2\) and \(U = 2E_2\).
\(E_2 = -3.4\,\text{eV}\), \(K = 3.4\,\text{eV}\), \(U = -6.8\,\text{eV}\).
On \(n = 2\) the electron has \(3.4\,\text{eV}\) of kinetic energy and \(-6.8\,\text{eV}\) of potential energy, a total of \(-3.4\,\text{eV}\). It is bound by \(3.4\,\text{eV}\).
Why is the total energy of the electron on a Bohr orbit negative?
Find the kinetic energy, in eV, of the electron of the helium ion \(\text{He}^+\) (\(Z = 2\)) on the orbit \(n = 3\).
An electron of hydrogen moves to an orbit of larger \(n\). Which of these increase?
How long one turn takes
The electron goes once round its orbit in the time \(T = 2\pi r/v\). The radius grows as \(n^2/Z\) and the speed falls as \(Z/n\), so the period grows as \(n^3/Z^2\). For hydrogen's lowest orbit, \(T_1 = 2\pi \times 0.529 \times 10^{-10}/2.18 \times 10^6 = 1.52 \times 10^{-16}\,\text{s}\).
The frequency of revolution is \(f = 1/T\), so \(f_1 = 6.56 \times 10^{15}\) turns per second. The electron on \(n = 2\) takes \(2^3 = 8\) times as long, and on \(n = 3\) it takes \(27\) times as long. That is the ratio you watched in Fig 3.1.
The electron as a current
A circling electron carries charge round a loop, so it is a tiny electric current. A current is the charge passing a point per second. The electron passes any point of its orbit once in each period, carrying charge \(e\), so the current is \(I = e/T\). On hydrogen's lowest orbit this is \(1.6 \times 10^{-19}/1.52 \times 10^{-16} = 1.05 \times 10^{-3}\,\text{A}\), about a milliampere.
For the electron of hydrogen, find three things. (a) The period and the frequency of revolution on the orbit \(n = 1\). (b) The period on \(n = 2\). (c) The current on \(n = 2\).
(a) \(T_1 = 2\pi r_1/v_1 = 2\pi \times 0.529 \times 10^{-10}/(2.18 \times 10^6) = 1.52 \times 10^{-16}\,\text{s}\), so \(f_1 = 1/T_1 = 6.56 \times 10^{15}\,\text{Hz}\).
(b) \(T \propto n^3\), so \(T_2 = 8T_1 = 1.22 \times 10^{-15}\,\text{s}\).
(c) \(I_2 = e/T_2 = 1.6 \times 10^{-19}/1.22 \times 10^{-15} = 1.31 \times 10^{-4}\,\text{A}\), one eighth of the current on \(n = 1\).
What is the ratio of the period of revolution on the orbit \(n = 3\) of hydrogen to the period on \(n = 1\)?
How does the period on the orbit \(n = 1\) of \(\text{He}^+\) compare with the period on \(n = 1\) of hydrogen?
A current loop acts as a small magnet, with a magnetic moment equal to the current times the area of the loop. For a Bohr orbit this works out to \(\mu = n\,eh/4\pi m\), about \(n \times 9.3 \times 10^{-24}\,\text{J/T}\). The nuclear charge cancels. The unit \(eh/4\pi m\) is the Bohr magneton, the natural size of an atom's magnetism.
Now return to the radio signals received at Ooty. Their atoms sit on orbits with \(n\) near 272. Treat one as hydrogen-like: the radius is \(0.529\,\text{Å} \times 272^2 = 3.9 \times 10^{-6}\,\text{m}\), about \(4\,\text{μm}\), as large as a bacterium. A jump from 272 to 271 releases so little energy that the photon is a radio wave, near \(328\,\text{MHz}\).
- Using \(r \propto n\) or \(v \propto 1/n^2\).\(r \propto n^2/Z\) and \(v \propto Z/n\). The energy, not the speed, goes as \(1/n^2\).
- Writing the kinetic energy as negative.\(K = -E\) is positive, since \(E\) is negative. \(U = 2E\) is the negative one.
- Forgetting that energies go as \(Z^2\).One \(Z\) comes from the charge, one from the smaller orbit: \(E_n = -13.6\,Z^2/n^2\) eV.