Bohr–Sommerfeld Theory and the Hydrogen Atom
The general Bohr–Sommerfeld quantization condition, and how to apply it to any periodic system from its phase diagram.
Bohr's model of hydrogen: quantized radius, speed, and energy.
Scaling with atomic number \(Z\) and with reduced mass \(\mred\) — the source of most numerical questions in this chapter.
The hydrogen spectrum: Rydberg constant, spectral series, isotope shift.
Sommerfeld's elliptical orbits, degeneracy, and the first appearance of the fine structure constant \(\alpha\).
What the model gets right, and precisely where it fails.
1.1 The Bohr–Sommerfeld Quantization Condition
For any physical system whose coordinates are periodic functions of time, there exists one quantum condition for each coordinate: \begin{equation} \oint p_i \,\dd q_i = n_i h , \qquad n_i = 1,2,3,\dots \label{eq:bs-rule} \end{equation} Here \(q_i\) is a generalized coordinate, \(p_i\) is the momentum conjugate to it, \(n_i\) is the quantum number for that coordinate, and \(\oint\) means the integral is taken over one complete period of \(q_i\). The constant \(h = 6.626\times 10^{-34}\,\mathrm{J}\,\mathrm{s}\) is Planck's constant.
A plot of \(p\) (vertical axis) against \(q\) (horizontal axis) is called the phase diagram of the motion. In classical mechanics the area enclosed by the phase trajectory may take any value whatever. Equation \(\eqref{eq:bs-rule}\) says something much stronger:
The area enclosed by the phase trajectory is not continuous — it is quantized in units of \(h\): \[ \text{Area} = \oint p\,\dd q = nh . \] This one sentence is the whole of old quantum theory. Every worked example below is an application of it.
A particle of mass \(m\) is confined to a one-dimensional box of width \(L\). The particle is free inside the box. Using Bohr–Sommerfeld theory, find the quantized energy of the particle.
Solution. Inside the box the particle is free, so its momentum has constant magnitude \(p\); it travels from one wall to the other with momentum \(+p\) and returns with \(-p\). One complete period therefore covers a path length \(2L\): \[ \oint p\,\dd x = p\,L + p\,L = 2pL = nh \quad\Longrightarrow\quad p = \frac{nh}{2L}. \] Hence \begin{equation} E = \frac{p^{2}}{2m} = \frac{n^{2}h^{2}}{8mL^{2}}, \qquad n = 1,2,3,\dots \label{eq:box-energy} \end{equation} The energy is quantized as \(E \propto n^{2}\).
This is the exact result of the Schrödinger equation for an infinite square well. The old quantum theory happens to be right here — but that is a coincidence of the flat potential, not a general feature, as the next example shows.
The classical energy of a harmonic oscillator is \(E = \dfrac{p^{2}}{2m} + \dfrac{1}{2}m\omega^{2}x^{2}\). Using Bohr–Sommerfeld theory, find the possible quantized energies.
Solution. Write the phase trajectory in standard form: \[ \frac{p^{2}}{2mE} + \frac{x^{2}}{\,2E/m\omega^{2}\,} = 1 , \] an ellipse with semi-axes \(p_{\max} = \sqrt{2mE}\) and \(x_{\max} = \sqrt{2E/m\omega^{2}}\).
The enclosed area of an ellipse is \(\pi\,(\text{semi-axis})(\text{semi-axis})\): \[ \oint p\,\dd x = \pi\, p_{\max}\, x_{\max} = \pi \sqrt{2mE}\,\sqrt{\frac{2E}{m\omega^{2}}} = \frac{2\pi E}{\omega}. \] Setting this equal to \(nh\), \begin{equation} \frac{2\pi E}{\omega} = nh \quad\Longrightarrow\quad \boxed{\,E = n\hbar\omega\,}, \qquad n = 1,2,3,\dots \label{eq:sho-energy} \end{equation}
Compare with the exact quantum result \(E = \left(n+\tfrac12\right)\hbar\omega\). Bohr–Sommerfeld reproduces the spacing \(\hbar\omega\) correctly but misses the zero-point energy \(\tfrac12\hbar\omega\) entirely. Questions often test exactly this discrepancy.
If you prefer to do the integral explicitly rather than quote the area of an ellipse, substitute \(x = x_{\max}\sin\theta\), so that \(\dd x = x_{\max}\cos\theta\,\dd\theta\) and \(p = p_{\max}\cos\theta\). The limits \(x = -x_{\max} \to +x_{\max}\) become \(\theta = -\pi/2 \to +\pi/2\), and \[ \oint p\,\dd x = 2 p_{\max} x_{\max}\!\!\int_{-\pi/2}^{\pi/2}\!\cos^{2}\theta\,\dd\theta = 2 p_{\max} x_{\max}\cdot\frac{\pi}{2} = \pi p_{\max} x_{\max}, \] as before.
The classical energy of a system is \(E = \dfrac{p^{2}}{2m} + \lambda\lvert x\rvert\).
For a given energy \(E\), the area enclosed by the phase diagram is proportional to \(E^{s}\). Find \(s\).
Using Bohr–Sommerfeld theory the quantized energy is proportional to \(n^{t}\), where \(n\) is the quantum number. Find \(t\).
Solution. (a) The turning points are where \(p=0\), i.e. \(\lvert x\rvert = E/\lambda\). By the symmetry of the trajectory about both axes it is enough to compute one quadrant and multiply by four: \[ \oint p \,\dd x = 4\!\int_{0}^{E/\lambda}\!\!\sqrt{2m\!\left(E - \lambda x\right)}\;\dd x . \] Put \(u = E - \lambda x\), so \(\dd x = -\dd u/\lambda\); the limits run \(u = E \to 0\): \[ \oint p\,\dd x = \frac{4\sqrt{2m}}{\lambda}\int_{0}^{E}\! u^{1/2}\,\dd u = \frac{8\sqrt{2m}}{3\lambda}\;E^{3/2}. \] So the area \(\propto E^{3/2}\) and \[ \boxed{s = \tfrac{3}{2}}. \]
(b) Setting the area equal to \(nh\), \[ \frac{8\sqrt{2m}}{3\lambda}E^{3/2} = nh \quad\Longrightarrow\quad E = \left(\frac{3\lambda h}{8\sqrt{2m}}\right)^{2/3} n^{2/3} \;\propto\; n^{2/3}, \] so \[ \boxed{t = \tfrac{2}{3}}. \]
Learn the general result. For \(V = \lambda\lvert x\rvert^{\,k}\), \[ \oint p\,\dd x \propto E^{\frac{1}{2}+\frac{1}{k}} \qquad\Longrightarrow\qquad E \propto n^{\frac{2k}{k+2}} . \] Check it: \(k=2\) (oscillator) gives \(E\propto n\); \(k=1\) gives \(E\propto n^{2/3}\); \(k\to\infty\) (box) gives \(E\propto n^{2}\). All three examples above in one line.
1.2 Bohr's Model of the Hydrogen Atom
A hydrogen atom is the neutral atom of the element hydrogen: a single positively charged proton with a single electron bound to it by the Coulomb force. Bohr's model treats the electron as moving in a stable circular orbit about the nucleus.
Bohr's boldest assumption is that a definite circular orbit is stable — he never explained why an accelerating charge on such an orbit does not radiate away its energy. The model is called semi-classical because it computes the orbit by classical mechanics and then imposes a quantum condition on the classical angular momentum by hand. The stability of the electronic state was only explained later by the Schrödinger equation.
1.2.1 The assumptions (Bohr's postulates)
In 1913 Niels Bohr proposed a model that reproduced the measured hydrogen spectrum to remarkable accuracy, at the price of one additional assumption about the Coulomb attraction between the electron and the nucleus. The whole model rests on four postulates:
Classical orbits. An atomic electron travels in a circular orbit around the nucleus, held there by the Coulomb attraction, and while it does so it obeys the ordinary laws of classical mechanics.
Quantized angular momentum. Of the continuous infinity of orbits that classical mechanics would allow, only those are realized for which the orbital angular momentum \(L\) is a whole-number multiple of \(\hbar = h/2\pi\): \[ L = n\hbar , \qquad n = 1,2,3,\dots \]
Non-radiating stationary states. An electron in one of these permitted orbits emits no electromagnetic radiation, even though it is being accelerated continuously. Its total energy \(E\) therefore stays fixed, and the orbit is called a stationary state.
Radiation on transition (Bohr frequency condition). Radiation appears only when the electron jumps discontinuously from a stationary state of energy \(E_i\) to one of energy \(E_f\). The frequency \(\nu\) of the light emitted is fixed by the energy difference divided by Planck's constant: \[ \nu = \frac{E_i - E_f}{h}. \]
Note carefully what each postulate does. P1 imports classical mechanics; P2 is the quantization that selects the allowed orbits; P3 rescues the atom from the classical instability that would otherwise make it collapse; P4 connects the level scheme to the observed spectrum. Questions that ask “which postulate is violated by classical electrodynamics?” are asking about P3.
The second postulate introduces the quantization of orbital angular momentum; the fourth introduces the quantization of the emitted radiation. Postulates P1 and P3 are, respectively, an inherited assumption and a prohibition — neither is derived from anything. This is precisely why the model is called ad hoc, and why the Schrödinger equation was needed to replace it.
1.2.2 The quantization rule
Take the motion to lie in a plane, so the natural generalized coordinate is the azimuthal angle \(\phi\), running from \(0\) to \(2\pi\) over one period. The conjugate momentum \(p_\phi\) is the angular momentum \(L\), which is constant for a central force. Applying \(\eqref{eq:bs-rule}\): \[ \oint p_{\phi}\,\dd\phi = L\!\int_{0}^{2\pi}\!\dd\phi = 2\pi L = nh , \] giving Bohr's celebrated quantization rule \begin{equation} L = m v r = \frac{nh}{2\pi} = n\hbar , \qquad n = 1,2,3,\dots \label{eq:bohr-quantization} \end{equation}
1.2.3 The de Broglie interpretation
Rewriting \(\eqref{eq:bohr-quantization}\) using \(p = mv\), \[ p\,(2\pi r) = nh \quad\Longrightarrow\quad 2\pi r = n\frac{h}{p} = n\lambda , \] where \(\lambda = h/p\) is the de Broglie wavelength. So \begin{equation} 2\pi r_{n} = n \lambda_{n} . \label{eq:debroglie-orbit} \end{equation}
The circumference of an allowed Bohr orbit is an integer number of de Broglie wavelengths — the orbit carries a standing electron wave. This is the physical content of Bohr's otherwise arbitrary rule, and it is a standard one-line question.
1.2.4 Radius, speed and energy
The electron is held in its circular orbit by electrostatic attraction, so the centripetal force equals the Coulomb force: \begin{equation} \frac{m v^{2}}{r} = \kc\,\frac{e^{2}}{r^{2}} , \label{eq:force-balance} \end{equation} where \(m\) and \(e\) are the electron's mass and charge, and \(\kc = 8.99\times 10^{9}\,\mathrm{N}\,\mathrm{m}^{2}\,\mathrm{C}^{-2}\) is Coulomb's constant. Hence the speed at any radius is \begin{equation} v = \sqrt{\,\ke\,\frac{1}{mr}\,} . \label{eq:speed-r} \end{equation}
The total energy at radius \(r\) is kinetic plus potential: \[ E = \frac{1}{2}mv^{2} - \ke\frac{1}{r} . \] Substituting \(mv^{2} = \ke\dfrac{1}{r}\) from \(\eqref{eq:force-balance}\), \begin{equation} E = \frac{1}{2}\ke\frac{1}{r} - \ke\frac{1}{r} = -\frac{1}{2}\,\ke\,\frac{1}{r} . \label{eq:E-of-r} \end{equation}
The total energy is exactly one half of the potential energy, and equal in magnitude to the kinetic energy with opposite sign. This is the virial theorem for a \(1/r\) potential — see \(\eqref{eq:virial}\).
Now impose quantization. From \(\eqref{eq:bohr-quantization}\), \(v = n\hbar/mr\); substituting into \(\eqref{eq:force-balance}\) and solving for \(r\): \begin{equation} \boxed{\;r_{n} = \frac{4\pi\varepsilon_{0}\hbar^{2}}{m e^{2}}\,n^{2} = n^{2}\abohr \;} \label{eq:bohr-radius} \end{equation} where \begin{equation} \abohr = \frac{4\pi\varepsilon_{0}\hbar^{2}}{m e^{2}} = 0.529\,\mathrm{Å} = 0.529\times 10^{-10}\,\mathrm{m} \label{eq:a0} \end{equation} is the Bohr radius. The speed is likewise quantized: \begin{equation} \boxed{\;v_{n} = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{n\hbar} = \frac{\alpha c}{n}\;} \label{eq:bohr-speed} \end{equation} with \(\alpha \approx 1/137\) the fine structure constant (see 1.6).
Finally, putting \(\eqref{eq:bohr-radius}\) into \(\eqref{eq:E-of-r}\): \begin{equation} \boxed{\;E_{n} = -\frac{m e^{4}}{2(4\pi\varepsilon_{0})^{2}\hbar^{2}} \frac{1}{n^{2}} = -\frac{\Ryd}{n^{2}} = -\frac{13.6\,\mathrm{eV}}{n^{2}}\;} \label{eq:bohr-energy} \end{equation} where the Rydberg energy is \begin{equation} \Ryd = \frac{m e^{4}}{2(4\pi\varepsilon_{0})^{2}\hbar^{2}} = \frac{1}{2}m c^{2}\alpha^{2} = 13.6\,\mathrm{eV} . \label{eq:rydberg-energy} \end{equation}
1.3 Hydrogen-like Atoms and the \(Z\) Scaling
A hydrogen-like (or hydrogenic) atom or ion is any nucleus of charge \(+Ze\) with exactly one bound electron. Because only two particles interact, and through a potential that depends only on their separation, every result of 1.2 carries over with the single replacement \(e^{2} \to Z e^{2}\) in the Coulomb attraction. Familiar examples are \(\mathrm{H}\) itself, \(\mathrm{He}^{+}\), \(\mathrm{Li}^{2+}\), \(\mathrm{Be}^{3+}\), and in general an ion carrying charge \(+(Z-1)e\).
Repeating the derivation with \(V(r) = -Ze^{2}/4\pi\varepsilon_{0}r\): \begin{align} r_{n} &= \frac{n^{2}}{Z}\,\abohr , \label{eq:r-Z}\\[2pt] v_{n} &= \frac{Z}{n}\,\alpha c , \label{eq:v-Z}\\[2pt] E_{n} &= -\,\frac{Z^{2}}{n^{2}}\;13.6\,\mathrm{eV} . \label{eq:E-Z} \end{align}
| Quantity | Symbol | Scaling | Value at \(n=1\), \(Z=1\) |
|---|---|---|---|
| Orbit radius | \(r_n\) | \(n^{2}/Z\) | \(0.529\,\mathrm{Å}\) |
| Speed | \(v_n\) | \(Z/n\) | \(\alpha c = 2.19\times 10^{6}\,\mathrm{m}\,\mathrm{s}^{-1}\) |
| Angular momentum | \(L\) | \(n\) | \(\hbar\) |
| Total energy | \(E_n\) | \(Z^{2}/n^{2}\) | \(-13.6\,\mathrm{eV}\) |
| Time period | \(T_n\) | \(n^{3}/Z^{2}\) | \(1.52\times 10^{-16}\,\mathrm{s}\) |
| Angular frequency | \(\omega_n\) | \(Z^{2}/n^{3}\) | \(4.13\times 10^{16}\,\mathrm{rad}\,\mathrm{s}^{-1}\) |
| Orbital current | \(i_n\) | \(Z^{2}/n^{3}\) | \(1.05\times 10^{-3}\,\mathrm{A}\) |
| Magnetic field at nucleus | \(B_n\) | \(Z^{3}/n^{5}\) | \(\sim12.5\,\mathrm{T}\) |
| Magnetic moment | \(\mu_n\) | \(n\) | \(\mu_{\mathrm B}\) |
1.3.1 Energies, and the virial theorem
1.3.2 The virial theorem.
For any system executing bounded motion, the time-averaged kinetic energy \(\langle T\rangle\) is related to the forces acting by \begin{equation} 2\,\langle T\rangle = -\Big\langle \textstyle\sum_i \vec F_i\cdot\vec r_i \Big\rangle . \label{eq:virial-general} \end{equation} If the potential is a homogeneous power law, \(V(r) = \lambda r^{\,k}\), this reduces to the form worth memorizing: \begin{equation} \boxed{\;2\,\langle T\rangle = k\,\langle V\rangle\;} \label{eq:virial-powerlaw} \end{equation} For the Coulomb potential \(V(r) = -\ke\,\dfrac{1}{r}\) we have \(k=-1\), so \begin{equation} 2\langle T\rangle = -\langle V\rangle, \qquad \langle T\rangle = -\tfrac12\langle V\rangle, \qquad E = \langle T\rangle + \langle V\rangle = \tfrac12\langle V\rangle = -\langle T\rangle . \label{eq:virial-coulomb} \end{equation}
Applying \(\eqref{eq:virial-coulomb}\) to every hydrogenic state, \begin{equation} T = -E_{n} = +\frac{Z^{2}}{n^{2}}\,13.6\,\mathrm{eV}, \qquad V = 2E_{n} = -\frac{2Z^{2}}{n^{2}}\,13.6\,\mathrm{eV}, \qquad \frac{T}{\lvert V\rvert} = \frac{1}{2}. \label{eq:virial} \end{equation}
Check \(\eqref{eq:virial-powerlaw}\) against the harmonic oscillator: \(k=2\) gives \(\langle T\rangle = \langle V\rangle\), i.e.\ the energy is shared equally — the familiar result. For the Coulomb case, \(k=-1\), the potential energy is twice the total energy and the kinetic energy is its negative. Given any one of \(T\), \(V\), \(E\) for a hydrogenic state you can write down the other two immediately.
1.3.3 Ionization energy
If the electron starts in the ground state \(n_i=1\), the minimum energy needed to free it is the energy required to reach the continuum edge, \(n_f = \infty\), where \(E_\infty = 0\): \begin{equation} E_{\text{ion}} = E_{\infty} - E_{1} = 0 - \left(-\frac{Z^{2}\Ryd}{1}\right) = Z^{2}\,13.6\,\mathrm{eV}. \label{eq:ionization} \end{equation} For hydrogen this is \(13.6\,\mathrm{eV}\); for \(\mathrm{He}^+\), \(54.4\,\mathrm{eV}\); for \(\mathrm{Li}^{2+}\), \(122.4\,\mathrm{eV}\).
1.4 Correction for Finite Nuclear Mass
Bohr assumed an infinitely heavy, motionless nucleus. If instead the nucleus has finite mass \(M\), both particles orbit their common centre of mass and the problem is a genuine two-body problem. It reduces to an equivalent one-body problem by replacing the electron mass \(m\) with the reduced mass \begin{equation} \mred = \frac{mM}{m+M} = \frac{m}{1 + m/M} . \label{eq:reduced-mass} \end{equation}
When \(M \gg m\) we have \(\mred \to m\) and the correction vanishes — which is why Bohr's infinite-mass assumption gave results so close to experiment for hydrogen (\(M_p/m \approx 1836\), so the correction is about \(0.05\%\)). The correction becomes decisive when the two masses are comparable (positronium) or when the light particle is heavy (muonic atoms).
Making the replacement \(m \to \mred\) throughout \(\eqref{eq:r-Z}\)–\(\eqref{eq:E-Z}\): \begin{align} r_{n} &= \frac{4\pi\varepsilon_{0}\hbar^{2}}{\mred e^{2}}\frac{n^{2}}{Z} = \frac{m}{\mred}\cdot\frac{n^{2}\abohr}{Z} &&\text{(radius scales as } 1/\mred), \label{eq:r-mu}\\[2pt] v_{n} &= \frac{Z}{n}\alpha c &&\text{(\textbf{no} reduced-mass correction)}, \label{eq:v-mu}\\[2pt] E_{n} &= -\frac{\mred}{m}\,\frac{Z^{2}}{n^{2}}\,13.6\,\mathrm{eV} &&\text{(energy scales as } \mred). \label{eq:E-mu} \end{align}
Three facts win most of the marks in this section: \(r \propto 1/\mred\), \ \(E \propto \mred\), \ and \(v\) is independent of \(\mred\). The last one is the least known and the most often tested.
| System | Constituents | \(\mred\) | Ground state energy |
|---|---|---|---|
| Hydrogen | \(p^{+}\,\electron\) | \(0.99946\,m_e\) | \(-13.6\,\mathrm{eV}\) |
| Deuterium | \(d^{+}\,\electron\) | \(0.99973\,m_e\) | \(-13.6\,\mathrm{eV}\) (shifted) |
| Positronium | \(\positron\,\electron\) | \(m_e/2\) | \(-6.8\,\mathrm{eV}\) |
| Muonium | \(\mu^{+}\,\electron\) | \(0.995\,m_e\) | \(\approx-13.5\,\mathrm{eV}\) |
| Muonic hydrogen | \(p^{+}\,\muminus\) | \(186\,m_e\) | \(\approx-2.53\,\mathrm{keV}\) |
| Muonic atom, charge \(Z\) | nucleus \(\muminus\) | \(\dfrac{207\,m_e M}{207 m_e + M}\) | \(-\,\dfrac{\mred}{m_e}Z^{2}\,13.6\,\mathrm{eV}\) |
1.4.1 Isotope effect
Examined at high resolution, each hydrogen spectral line is accompanied by a faint companion line. These extra lines come from the heavy isotope deuterium (\(^{2}_{1}\mathrm{H}\), or \(d\)) present in ordinary hydrogen gas at about one part in \(6700\). Since \(M_d \approx 2M_p\), deuterium has a slightly larger reduced mass than hydrogen, hence slightly larger transition energies and slightly shorter wavelengths.
The Balmer-\(\alpha\) line of hydrogen lies at \(6563\,\mathrm{Å}\). Estimate the wavelength separation between the hydrogen and deuterium components. Take \(M_p = 1836\,m_e\) and \(M_d = 3672\,m_e\).
Solution. Since \(E \propto \mred\) and \(\lambda \propto 1/E\), \[ \frac{\lambda_{\mathrm H}}{\lambda_{\mathrm D}} = \frac{\mred^{\mathrm D}}{\mred^{\mathrm H}} = \frac{M_d}{1+M_d/m_e}\cdot\frac{1+M_p/m_e}{M_p} = \frac{3672}{3673}\cdot\frac{1837}{1836} . \] Using \(3672 = 2\times 1836\), \[ \frac{\lambda_{\mathrm H}}{\lambda_{\mathrm D}} = \frac{2\times 1837}{3673} = \frac{3674}{3673} = 1.000272 . \] Therefore \[ \Delta\lambda = \lambda_{\mathrm H} - \lambda_{\mathrm D} \approx \lambda_{\mathrm H}\times 2.72\times 10^{-4} \approx 1.8\,\mathrm{Å}, \] in excellent agreement with the observed separation. Urey's discovery of deuterium in 1932 rested on precisely this measurement.
1.5 The Spectrum of the Hydrogen Atom
An electron may pass from one stationary state to another by absorbing energy and moving up, or by giving off energy and moving down. The atom absorbs or releases this energy as electromagnetic radiation, of energy equal to the difference between the two levels: \begin{equation} \Delta E = E_{n_f} - E_{n_i} = -Z^{2}\,\Ryd\!\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right). \label{eq:transition-energy} \end{equation} If \(n_f > n_i\) then \(\Delta E > 0\) and the atom absorbs radiation; if \(n_f < n_i\) then \(\Delta E < 0\) and the atom emits radiation.
1.5.1 Wavelength and the Rydberg constant
Using the Planck relation \(\lvert\Delta E\rvert = hc/\lambda\), \begin{equation} \frac{1}{\lambda} = R\,Z^{2}\!\left(\frac{1}{n_f^{2}} - \frac{1}{n_i^{2}}\right), \qquad n_i > n_f , \label{eq:rydberg-formula} \end{equation} where the Rydberg constant is \begin{equation} R = \frac{\mred e^{4}}{8\varepsilon_{0}^{2}h^{3}c} = \frac{\mred}{m}\,\Ry , \qquad \Ry = 1.097\times 10^{7}\,\mathrm{m}^{-1} . \label{eq:rydberg-constant} \end{equation} \(\Ry\) is the value for an infinitely heavy nucleus; for a real hydrogen-like atom of reduced mass \(\mred\) use the first form.
1.5.2 Spectral series
When an electric discharge is passed through hydrogen gas the \(\mathrm{H_2}\) molecules dissociate, and the energetically excited \(\mathrm H\) atoms emit light at discrete frequencies, producing a spectrum of lines grouped into series by their common lower level \(n_f\).
| Series | \(n_f\) | \(n_i\) | Region | Series limit |
|---|---|---|---|---|
| Lyman | 1 | \(2,3,4,\dots\) | Ultraviolet | \(911.8\,\mathrm{Å}\) |
| Balmer | 2 | \(3,4,5,\dots\) | Visible | \(3646\,\mathrm{Å}\) |
| Paschen | 3 | \(4,5,6,\dots\) | Near infrared | \(8206\,\mathrm{Å}\) |
| Brackett | 4 | \(5,6,7,\dots\) | Infrared | \(1.459\,\mathrm{\mu m}\) |
| Pfund | 5 | \(6,7,8,\dots\) | Far infrared | \(2.279\,\mathrm{\mu m}\) |
| Humphreys | 6 | \(7,8,9,\dots\) | Far infrared | \(3.282\,\mathrm{\mu m}\) |
The longest wavelength (smallest energy) in a series is the transition from \(n_i = n_f+1\); the shortest is the series limit \(n_i \to \infty\), at \(1/\lambda = R Z^{2}/n_f^{2}\). For the Balmer series the first line (\(3\to2\)) is H\(_\alpha\) at \(6563\,\mathrm{Å}\) and the second (\(4\to2\)) is H\(_\beta\) at \(4861\,\mathrm{Å}\).
1.5.3 Number of spectral lines
If an atom is excited to level \(n\) and then cascades down by all possible routes, the number of distinct emission lines is the number of ways of choosing an ordered pair of levels from \(n\) available levels: \begin{equation} N = \binom{n}{2} = \frac{n(n-1)}{2}. \label{eq:num-lines} \end{equation} More generally, for a cascade confined between levels \(n_f\) and \(n_i\), \begin{equation} N = \frac{(n_i - n_f)(n_i - n_f + 1)}{2}. \label{eq:num-lines-general} \end{equation}
A hydrogen atom at rest emits a photon in going from the \(n=2\) state to the \(n=1\) state. Find the recoil speed of the atom after the transition.
Solution. The transition energy is \[ \Delta E = 13.6\left(1 - \tfrac14\right) = 10.2\,\mathrm{eV}. \] Linear momentum is conserved, and the atom was initially at rest, so the atom's recoil momentum equals the photon momentum: \[ M v = \frac{\Delta E}{c} \quad\Longrightarrow\quad v = \frac{\Delta E}{M c}. \] With \(M = 1.67\times 10^{-27}\,\mathrm{kg}\) and \(\Delta E = 10.2 \times 1.6\times 10^{-19}\,\mathrm{J} = 1.63\times 10^{-18}\,\mathrm{J}\), \[ v = \frac{1.63\times10^{-18}}{1.67\times10^{-27}\times 3\times10^{8}} \approx 3.3\,\mathrm{m}\,\mathrm{s}^{-1}. \]
The recoil energy \(\tfrac12 M v^{2} \sim 5.6\times 10^{-9}\,\mathrm{eV}\) is utterly negligible next to \(10.2\,\mathrm{eV}\) — which justifies ignoring recoil in Eq. \(\eqref{eq:transition-energy}\). It is not negligible in nuclear \(\gamma\)-emission, which is why the Mössbauer effect exists.
A single electron orbits a stationary nucleus of charge \(+Ze\). It requires \(47.2\,\mathrm{eV}\) to excite the electron from the second to the third Bohr orbit. Find
the value of \(Z\);
the energy required to excite the electron from the third to the fourth Bohr orbit;
the wavelength of radiation needed to remove the electron from the first Bohr orbit to infinity;
the kinetic energy, potential energy and angular momentum in the first Bohr orbit;
the radius of the first Bohr orbit.
Solution. (a) From \(\eqref{eq:E-Z}\), \[ \Delta E_{2\to3} = 13.6\,Z^{2}\!\left(\frac{1}{4}-\frac{1}{9}\right) = 13.6\,Z^{2}\times\frac{5}{36} = 47.2 \;\Longrightarrow\; Z^{2} = 25,\quad \boxed{Z = 5}. \]
(b) \[ \Delta E_{3\to4} = 13.6\times 25\left(\frac{1}{9}-\frac{1}{16}\right) = 340 \times \frac{7}{144} = 16.53\,\mathrm{eV}. \]
(c) To remove the electron from \(n=1\) to \(n=\infty\) requires \(E_{\text{ion}} = 13.6\times25 = 340\,\mathrm{eV}\). Hence \[ \lambda = \frac{hc}{E_{\text{ion}}} = \frac{12400\,\mathrm{eV}\,\mathrm{Å}}{340\,\mathrm{eV}} \approx 36.5\,\mathrm{Å}. \]
(d) \(E_{1} = -340\,\mathrm{eV}\), so by \(\eqref{eq:virial}\) \[ T = 340\,\mathrm{eV}, \qquad V = -680\,\mathrm{eV}, \qquad L = 1\cdot\hbar = 1.05\times 10^{-34}\,\mathrm{J}\,\mathrm{s}. \]
(e) From \(\eqref{eq:r-Z}\), \[ r_{1} = \frac{1^{2}}{5}\times0.529\,\mathrm{Å} = 0.106\,\mathrm{Å}. \]
\(\mathrm{Li}^{2+}\) is hydrogen-like with \(Z=3\).
Find the wavelength of the radiation required to excite the electron from the first to the third Bohr orbit.
How many spectral lines are observed in the emission spectrum of the excited system?
Solution. (a) With \(Z=3\), \(n_i=1\), \(n_f=3\): \[ \Delta E = 13.6\times 9\left(1 - \frac19\right) = 122.4\times \frac{8}{9} = 108.8\,\mathrm{eV}, \] \[ \lambda = \frac{12400\,\mathrm{eV}\,\mathrm{Å}}{108.8\,\mathrm{eV}} \approx 114\,\mathrm{Å}. \]
(b) From \(\eqref{eq:num-lines}\) with \(n=3\): \[ N = \frac{3\times 2}{2} = 3 , \] the transitions being \(3\to2\), \(3\to1\) and \(2\to1\).
A muonic atom consists of a nucleus of charge \(Ze\) and a negative muon \(\muminus\) orbiting it. The muon has charge \(-e\) and a mass \(207\) times the electron mass.
Calculate the muon–nucleus separation of the first Bohr orbit for \(Z=1\).
Calculate the binding energy of this muonic atom.
What is the wavelength of the first line of the Lyman series for such an atom?
Solution. (a) With \(m_\mu = 207\,m_e\) and \(M = M_p = 1836\,m_e\), the reduced mass from \(\eqref{eq:reduced-mass}\) is \[ \mred = \frac{207 \times 1836}{207 + 1836}\,m_e = \frac{380052}{2043}\,m_e = 186\,m_e . \] Then from \(\eqref{eq:r-mu}\) with \(n=1\), \(Z=1\), \[ r_{1} = \frac{m_e}{\mred}\,\abohr = \frac{0.529\,\mathrm{Å}}{186} = 2.84\times 10^{-3}\,\mathrm{Å} = 284\,\mathrm{fm}. \]
The muon orbits far closer to the nuclear surface than an electron does — only a few hundred femtometres out. This is exactly what makes muonic atoms useful: their spectra are sensitive to nuclear size and structure.
(b) From \(\eqref{eq:E-mu}\) with \(n=1\), \(Z=1\), \[ E_{1} = -186 \times 13.6\,\mathrm{eV} = -2.53\,\mathrm{keV}, \] so the binding energy is \(2.53\,\mathrm{keV}\).
(c) The first Lyman line is \(n_i=2 \to n_f=1\): \[ \Delta E = 186\times 13.6\left(1-\tfrac14\right) = 2530 \times 0.75 = 1.90\,\mathrm{keV}, \] \[ \lambda = \frac{12400\,\mathrm{eV}\,\mathrm{Å}}{1.90\times10^{3}\,\eV} \approx 6.5\,\mathrm{Å}. \] The Lyman lines of muonic hydrogen therefore lie in the X-ray region, and X-ray techniques are needed to study muonic-atom spectra.
According to Bohr's theory, the magnetic field at the centre of the nucleus due to the orbital motion of the electron in the \(n\)th Bohr orbit of hydrogen is proportional to
- (a)
\(n^{-5}\)
- (b)
\(n^{-3}\)
- (c)
\(n^{-4}\)
- (d)
\(n^{-2}\)
Solution. (a). The orbiting electron is a current loop of current \(i = e/T = ev/2\pi r\), producing at its centre \[ B = \frac{\mu_{0} i}{2r} = \frac{\mu_{0} e v}{4\pi r^{2}} . \] Using \(v \propto 1/n\) and \(r \propto n^{2}\), \[ B \propto \frac{1/n}{(n^{2})^{2}} = \frac{1}{n^{5}} . \] More generally \(B \propto Z^{3}/n^{5}\), as recorded in Table 1.1.
The energy gap between the \(n=1\) and \(n=2\) levels of hydrogen is \(E_{0}\). Consider a muonic carbon ion \(\mathrm{C}^{5+}\): a \(^{12}_{\ 6}\mathrm{C}\) nucleus orbited by a muon (\(q=-e\), \(m_\mu = 210\,m_e\)). Find the energy of the photon emitted in the \(n=3 \to n=2\) transition, in terms of \(E_{0}\).
Solution. For hydrogen, \[ E_{0} = 13.6\left(1 - \tfrac14\right) = 13.6\times\tfrac34 . \] For the muonic ion, take \(\mred \approx m_\mu = 210\,m_e\) (the carbon nucleus is \(\sim\!100\times\) heavier than the muon, so the correction is small) and \(Z=6\): \[ \Delta E = 210 \times 36 \times 13.6 \left(\frac{1}{4}-\frac{1}{9}\right) = 210\times36\times13.6\times\frac{5}{36} = 210\times 5 \times 13.6 . \] Dividing by \(E_{0} = 13.6\times\tfrac34\): \[ \frac{\Delta E}{E_{0}} = \frac{210\times5}{3/4} = 1400 \quad\Longrightarrow\quad \boxed{\Delta E = 1400\,E_{0}} . \]
1.6 Sommerfeld's Extension: Elliptical Orbits
In Bohr's model the electron moves only on circles. Sommerfeld generalized this: the electron revolves about the nucleus on an ellipse, with the nucleus at one focus. The model is exactly analogous to Kepler's description of a planet orbiting the Sun.
1.6.1 Two quantum numbers
Polar coordinates \((r,\phi)\) give two periodic coordinates, hence two quantum conditions from \(\eqref{eq:bs-rule}\): \begin{align} \oint p_{r}\,\dd r &= n_{r} h , \qquad n_{r} = 0,1,2,\dots \label{eq:radial-quant}\\ \oint p_{\phi}\,\dd\phi &= n_{\phi} h , \qquad n_{\phi} = 1,2,3,\dots \label{eq:azimuthal-quant} \end{align} Here \(n_{r}\) is the radial quantum number and \(n_{\phi}\) the azimuthal quantum number. Since \(p_\phi = L\) is constant, \(\eqref{eq:azimuthal-quant}\) immediately gives \(L = n_\phi \hbar\), exactly Bohr's rule with \(n \to n_\phi\).
Evaluating the radial integral for a Kepler ellipse of eccentricity \(\varepsilon\) gives the central result \begin{equation} n_{r} + n_{\phi} \equiv n , \label{eq:n-total} \end{equation} defining the principal quantum number \(n\).
Many texts write \(k\) for \(n_\phi\). In this book \(n_\phi\) is used throughout; \(k\) appears only where a previous-year question uses it.
Since \(n_{r} \ge 0\) and \(n_{\phi} \ge 1\), the allowed values are \begin{equation} n_{\phi} = 1, 2, 3, \dots, n . \label{eq:nphi-range} \end{equation}
\(n_\phi = 0\) is excluded. It would mean \(L=0\), i.e.\ a degenerate straight-line "orbit" passing through the nucleus, which is not an ellipse at all. Note that quantum mechanics does allow \(\ell=0\) (\(s\) states) — this exclusion is one of the model's known defects.
1.6.2 Energy and degeneracy
Carrying the quantization through, the energy comes out as \begin{equation} E_{n} = -\frac{\mred e^{4} Z^{2}} {2(4\pi\varepsilon_{0})^{2}\hbar^{2}\,(n_r+n_\phi)^{2}} = -\frac{Z^{2}}{n^{2}}\,13.6\,\mathrm{eV} . \label{eq:sommerfeld-energy} \end{equation}
This is the same as the Bohr result \(\eqref{eq:E-Z}\), and it is the key point of the whole section: the energy depends only on the sum \(n=n_r+n_\phi\), not on how that sum is partitioned. Every one of the \(n\) orbits with \(n_\phi = 1,\dots,n\) has the same energy. This is the first appearance of degeneracy in atomic physics.
1.6.3 Size and shape of the orbits
The ellipse has \begin{align} \text{semi-major axis:}\quad a &= \frac{n^{2}}{Z}\,\abohr , \label{eq:semi-major}\\ \text{semi-minor axis:}\quad b &= \frac{n\,n_{\phi}}{Z}\,\abohr = a\,\frac{n_{\phi}}{n} , \label{eq:semi-minor}\\ \text{eccentricity:}\quad \varepsilon &= \sqrt{1 - \left(\frac{n_\phi}{n}\right)^{2}} . \label{eq:eccentricity} \end{align} So the size is fixed by \(n\) alone, and the shape by the ratio \(n_\phi/n\):
\(n_\phi = n\) \ (\(n_r = 0\)): \(b=a\), \(\varepsilon=0\) — a circle of radius \(n^{2}\abohr/Z\), recovering Bohr's orbit.
\(n_\phi < n\): \(b
\(n_\phi = 1\): the most eccentric orbit allowed for that \(n\).
1.6.4 Relativistic correction and the fine structure constant
Sommerfeld's real triumph was to notice that on a highly eccentric orbit the electron moves fast near perihelion, so its mass varies over the orbit. Using the relativistic kinetic energy \[ T = \sqrt{p^{2}c^{2} + m^{2}c^{4}} - mc^{2} \approx \frac{p^{2}}{2m} - \frac{p^{4}}{8m^{3}c^{2}} + \cdots \] the degeneracy in \(n_\phi\) is lifted, and the energy becomes \begin{equation} \boxed{\; E_{n,n_\phi} = -\frac{Z^{2}\,13.6\,\mathrm{eV}}{n^{2}} \left[\,1 + \frac{\alpha^{2}Z^{2}}{n^{2}} \left(\frac{n}{n_\phi} - \frac{3}{4}\right)\right]\;} \label{eq:sommerfeld-fine} \end{equation} where \begin{equation} \alpha = \frac{1}{4\pi\varepsilon_{0}}\frac{e^{2}}{\hbar c} = \frac{v_{1}}{c} \approx \frac{1}{137} \label{eq:alpha} \end{equation} is the fine structure constant — the ratio of the electron's speed in the first Bohr orbit of hydrogen to the speed of light in vacuum. It is dimensionless.
The splitting therefore scales as \begin{equation} \Delta E_{\text{fs}} \;\propto\; Z^{4}\alpha^{2} \;\propto\; \frac{Z^4}{n^4}, \label{eq:fs-scaling} \end{equation} which is roughly \(10^{-5}\) of the gross level spacing for hydrogen — hence the name fine structure.
Three high-yield facts: \(\alpha\) is dimensionless; \(\alpha = v_1/c\); and fine structure splitting goes as \(Z^{4}\). The \(Z^{4}\) dependence is why fine structure is easy to observe in heavy hydrogenic ions and hard in hydrogen.
Remarkably, \(\eqref{eq:sommerfeld-fine}\) gives the correct magnitude of the hydrogen fine structure, and yet the model is wrong: the splitting is really caused jointly by the relativistic correction, spin–orbit coupling and the Darwin term, and it depends on the total angular momentum \(j\), not on \(n_\phi\). Sommerfeld got the right answer for the wrong reason. The correct treatment is the subject of Chapter 2.
1.7 Successes and Failures of the Model
What the model explains
Discrete energy levels and the stability of atoms.
The hydrogen spectrum quantitatively, including all series and the value of \(R\) from first principles.
Spectra of all one-electron ions via \(Z^{2}\) scaling.
The isotope shift, and the spectra of positronium, muonium and muonic atoms via the reduced mass.
The correct order of magnitude of the fine structure splitting.
Moseley's law for X-ray lines (Chapter 5).
Where it fails
Gives no line intensities and no transition rates.
Fails completely for any atom with two or more electrons, even helium.
Assigns \(L = n\hbar \ne 0\) to the ground state; experiment and quantum mechanics give \(\ell = 0\).
Has no electron spin, hence cannot explain the anomalous Zeeman effect or the sodium doublet (Chapters 3–4).
Selection rules must be imposed ad hoc; they are not derived.
A definite orbit with simultaneously known \(r\) and \(p\) violates the uncertainty principle.
Cannot account for the Lamb shift or hyperfine structure.
The failures above are not repairable by patching the orbits. They are resolved by abandoning orbits altogether in favour of the wavefunction \(\psi_{n\ell m}\) obtained from the Schrödinger equation, and then adding the relativistic and spin–orbit corrections perturbatively. That programme — and the correct fine structure formula in terms of \(j\) — is Chapter 2.
Formula Summary
Quantization \[ \oint p_i\,\dd q_i = n_i h , \qquad L = n\hbar , \qquad 2\pi r_n = n\lambda_n \] Hydrogenic atom (put \(\mred \to m\), \(Z \to 1\) for hydrogen) \[ r_n = \frac{m}{\mred}\frac{n^{2}}{Z}\abohr, \qquad v_n = \frac{Z}{n}\alpha c, \qquad E_n = -\frac{\mred}{m}\frac{Z^{2}}{n^{2}}(13.6\,\mathrm{eV}) \] \[ \abohr = 0.529\,\mathrm{Å}, \qquad \Ryd = 13.6\,\mathrm{eV} = \tfrac12 mc^{2}\alpha^{2}, \qquad \alpha = \frac{1}{137} \] Virial relations \[ T = -E_n, \qquad V = 2E_n, \qquad E_{\text{ion}} = Z^{2}\,13.6\,\mathrm{eV} \] Reduced mass \[ \mred = \frac{mM}{m+M}: \qquad r \propto \frac{1}{\mred}, \qquad E \propto \mred, \qquad v \ \text{independent of}\ \mred \] Spectrum \[ \frac{1}{\lambda} = R Z^{2}\!\left(\frac{1}{n_f^{2}}-\frac{1}{n_i^{2}}\right), \qquad R = \frac{\mred}{m}\Ry, \qquad \Ry = 1.097\times 10^{7}\,\mathrm{m}^{-1} \] \[ hc = 12400\,\mathrm{eV}\,\mathrm{Å}, \qquad N_{\text{lines}} = \frac{n(n-1)}{2} \] Sommerfeld \[ n = n_r + n_\phi, \qquad n_\phi = 1,\dots,n, \qquad \frac{b}{a} = \frac{n_\phi}{n}, \qquad \varepsilon = \sqrt{1-\left(\tfrac{n_\phi}{n}\right)^{2}} \] \[ E_{n,n_\phi} = -\frac{Z^{2}(13.6\,\mathrm{eV})}{n^{2}} \left[1+\frac{\alpha^{2}Z^{2}}{n^{2}}\left(\frac{n}{n_\phi}-\frac34\right)\right], \qquad \Delta E_{\text{fs}} \propto Z^{4}\alpha^{2} \] Power-law potential \(V=\lambda\lvert x\rvert^{k}\): \(\;E \propto n^{2k/(k+2)}\)
Previous Year Questions
The wavefunction of which orbital is spherically symmetric?
- (a)
\(3s\)
- (b)
\(3p_x\)
- (c)
\(3d_{xy}\)
- (d)
\(3d_{z^{2}}\)
Positronium is an atom made of an electron and a positron. Given that the Bohr radius for the ground state of the hydrogen atom is \(0.53\,\mathrm{Å}\), the Bohr radius for the ground state of positronium is \(\mathrm{Å}\) (up to two decimal places).
Consider a gas of hydrogen atoms in the atmosphere of the Sun, where the temperature is \(5800\,\mathrm{K}\). If a sample from this atmosphere contains \(10^{18}\) hydrogen atoms in the ground state, the number of hydrogen atoms in the first excited state is approximately \(A \times 10^{n}\), where \(A\) is an integer. Find the value of \(n\). (Boltzmann constant \(k_B = 8.6\times 10^{-5}\,\mathrm{eV}\,\mathrm{K}^{-1}\).)
Which one of the following is a dimensionless constant?
- (a)
Permittivity of free space
- (b)
Permeability of free space
- (c)
Bohr magneton
- (d)
Fine structure constant
The screened nuclear charge of a neutral helium atom is \(1.7e\), where \(e\) is the magnitude of the electronic charge. Assuming the Bohr model of the atom, for which the energy levels are \[ E_{n} = -\frac{Z^{2}}{2}\,\frac{1}{n^{2}}\ \text{atomic units} \] (\(Z\) being the atomic number), the first ionization potential of helium in atomic units is
- (a)
\(0.89\)
- (b)
\(1.78\)
- (c)
\(0.94\)
- (d)
\(3.16\)
Given that the ground state energy of the hydrogen atom is \(-13.6\,\mathrm{eV}\), the ground state energy of positronium (a bound state of an electron and a positron) is
- (a)
\(+6.8\,\mathrm{eV}\)
- (b)
\(-6.8\,\mathrm{eV}\)
- (c)
\(-13.6\,\mathrm{eV}\)
- (d)
\(-27.2\,\mathrm{eV}\)
A muon \(\muminus\) from cosmic rays is trapped by a proton to form a hydrogen-like atom. Given that the muon is approximately \(200\) times heavier than the electron, the longest wavelength of the spectral line in the analogue of the Lyman series of such an atom will be
- (a)
\(6563\,\mathrm{Å}\)
- (b)
\(6.6\,\mathrm{Å}\)
- (c)
\(31\,\mathrm{Å}\)
- (d)
\(1216\,\mathrm{Å}\)
A negative muon, of mass nearly \(207\) times that of an electron, replaces an electron in a \(\mathrm{Li}\) atom. The lowest ionization energy for the muonic \(\mathrm{Li}\) atom is approximately
- (a)
\(207\) times larger than that of normal \(\mathrm{Li}\)
- (b)
the same as that of normal \(\mathrm{Li}\)
- (c)
\(207\) times smaller than that of normal \(\mathrm{Li}\)
- (d)
the same as that of normal \(\mathrm{H}\)
An atom of mass \(M\) can be excited to a state of mass \(M + \Delta M\) by photon capture. The frequency of a photon which can cause this transition is
- (a)
\(\dfrac{\Delta M\, c^{2}}{h}\)
- (b)
\(\dfrac{\Delta M\, c^{2}}{h}\left(1+\dfrac{\Delta M}{M}\right)\)
- (c)
\(\dfrac{\Delta M\, c^{2}}{h}\left(1-\dfrac{\Delta M}{2M}\right)\)
- (d)
\(\dfrac{\Delta M\, c^{2}}{h}\left(1+\dfrac{\Delta M}{2M}\right)\)
The wavelength of the first Balmer line of hydrogen is \(\lambda\). The wavelength of the corresponding line for a hydrogenic atom with \(Z=2\) and nuclear mass \(4M_p\) is approximately
- (a)
\(\lambda/4\)
- (b)
\(\lambda/2\)
- (c)
\(2\lambda\)
- (d)
\(4\lambda\)
In the absorption spectrum of the H atom, the frequency of the transition from the ground state to the first excited state is \(\nu_{\mathrm H}\). The corresponding frequency for a bound state of a positively charged muon \(\mu^{+}\) and an electron is \(\nu_{\mu}\). Using \(m_{\mu} = e-28\,\mathrm{kg}\), \(m_{e} = e-30\,\mathrm{kg}\) and \(M_p \gg m_e, m_\mu\), the value of \((\nu_{\mu}-\nu_{\mathrm H})/\nu_{\mathrm H}\) is
- (a)
\(0.001\)
- (b)
\(-0.001\)
- (c)
\(-0.01\)
- (d)
\(0.01\)
The bond dissociation energy of a molecule is defined as the energy required to dissociate it. For H2 and H2+ the bond dissociation energies are \(4.478\,\mathrm{eV}\) and \(2.651\,\mathrm{eV}\) respectively. If the equilibrium bond lengths of H2 and H2+ are identical, the ionization potential of the hydrogen molecule is closest to
- (a)
\(15.427\,\mathrm{eV}\)
- (b)
\(11.773\,\mathrm{eV}\)
- (c)
\(20.729\,\mathrm{eV}\)
- (d)
\(6.471\,\mathrm{eV}\)
The binding energy of the hydrogen atom (electron bound to proton) is \(13.6\,\mathrm{eV}\). The binding energy of positronium (electron bound to positron) is
- (a)
\(13.6/2\) eV
- (b)
\(13.6/1810\) eV
- (c)
\(13.6\times1810\) eV
- (d)
\(13.6\times2\) eV
If the proton were ten times lighter, the ground state energy of the electron in a hydrogen atom would be
- (a)
less
- (b)
more
- (c)
the same
- (d)
less, more or equal depending on the electron mass
A hydrogen atom in its ground state is collided with an electron of kinetic energy \(13.377\,\mathrm{eV}\). The maximum factor by which the radius of the atom would increase is
- (a)
\(7\)
- (b)
\(8\)
- (c)
\(49\)
- (d)
\(64\)
If the Rydberg constant of an atom of finite nuclear mass is \(R = \Ry\left(1 - 2.5\times 10^{-4}\right)\), where \(\Ry\) is the Rydberg constant for infinite nuclear mass, the ratio of the electronic to nuclear mass of the atom is approximately
- (a)
\(2.5\times 10^{-4}\)
- (b)
\(5\times 10^{-4}\)
- (c)
\(1.25\times 10^{-4}\)
- (d)
\(2.5\times 10^{-2}\)
Solutions to Previous Year Questions
Solution. Only \(s\) orbitals have \(\ell = 0\), for which the angular part is the constant \(Y_{0}^{0} = 1/\sqrt{4\pi}\). Hence \(3s\) is spherically symmetric; \(p\) and \(d\) orbitals have angular nodes.
Solution. Positronium has \(\mred = m_e/2\), and \(r \propto 1/\mred\): \[ a_{\text{Ps}} = \frac{m_e}{\mred}\abohr = 2\abohr = 2 \times 0.53 = 1.06\,\mathrm{Å}. \]
Solution. By the Boltzmann distribution, with degeneracies \(g_1=2\), \(g_2=8\), \[ \frac{N_2}{N_1} = \frac{g_2}{g_1}e^{-\Delta E/k_BT}, \qquad \Delta E = 10.2\,\mathrm{eV}. \] \[ \frac{\Delta E}{k_B T} = \frac{10.2}{8.6\times10^{-5}\times 5800} = \frac{10.2}{0.4988} = 20.45 . \] \[ N_2 = 10^{18}\times 4 \times e^{-20.45} = 10^{18}\times 4 \times 1.31\times10^{-9} \approx 5\times 10^{9}. \] Hmm — note that if degeneracy is ignored (\(g_2/g_1 = 1\)) one gets \(\approx 1\times10^{9}\). Either way the answer is reported as \(A\times10^{n}\) with \(n = 9\).
The original manuscript recorded the answer to this question as \(14\), which is inconsistent with \(T=5800\,\mathrm{K}\) and \(\Delta E=10.2\,\mathrm{eV}\). The GATE official key should be checked against the exact wording of the question before this goes to print — in particular whether the stated temperature is \(5800\,\mathrm{K}\) or \(58000\,\mathrm{K}\).
Solution. \(\alpha = e^{2}/4\pi\varepsilon_{0}\hbar c \approx 1/137\) is dimensionless. Permittivity, permeability and the Bohr magneton all carry units.
Solution. In neutral helium both electrons sit in \(n=1\), and each moves in the screened charge \(Z_{\text{eff}} = 1.7\): \[ E(\text{He}) = 2\times\left(-\frac{(1.7)^{2}}{2}\right) = -(1.7)^{2} = -2.89\ \text{a.u.} \] Removing one electron leaves He+, which is hydrogen-like. There is now no second electron to do any screening, so the full nuclear charge \(Z = 2\) acts: \[ E(\text{He}^{+}) = -\frac{2^{2}}{2} = -2\ \text{a.u.} \] The first ionization potential is the energy needed to go from one to the other: \[ I_{1} = E(\text{He}^{+}) - E(\text{He}) = (-2) - (-2.89) = \boxed{0.89\ \text{a.u.}} \]
The trap is carrying \(Z_{\text{eff}} = 1.7\) over to He+ as well, which gives \(-1.445\) and hence \(1.445\) — close to the distractor \(1.78\). Screening exists only when there is another electron to do the screening; a one-electron ion is always unscreened. As a sanity check, \(0.89\ \text{a.u.} = 0.89\times27.2\,\mathrm{eV} = 24.2\,\mathrm{eV}\), against the measured \(24.6\,\mathrm{eV}\).
Solution. For positronium \(\mred = m_e/2\), and \(E \propto \mred\): \[ E_1 = \frac{1}{2}\times(-13.6\,\mathrm{eV}) = -6.8\,\mathrm{eV}. \]
Solution. With \(m_\mu = 200\,m_e\) and \(M_p = 1836\,m_e\), \[ \mred = \frac{200\times1836}{2036}m_e \approx 180\,m_e . \] Energies scale by \(180\): \[ E_1 = -180\times13.6 = -2448\,\mathrm{eV}, \qquad E_2 = E_1/4 = -612\,\mathrm{eV}. \] The longest wavelength of the Lyman analogue is the smallest energy gap, \(n=2\to1\): \[ \Delta E = 2448 - 612 = 1836\,\mathrm{eV}, \qquad \lambda = \frac{12400\,\mathrm{eV}\,\mathrm{Å}}{1836} \approx 6.6\,\mathrm{Å}. \]
Solution. Ionization energy \(\propto \mred\). For the muonic Li atom the reduced mass of the muon–nucleus pair is \[ \mred = \frac{m_\mu M}{m_\mu + M} \approx m_\mu = 207\,m_e \] (since a Li nucleus is \(\sim\!7\times1836\,m_e \gg 207\,m_e\)). Hence \[ \frac{E_{\text{ion}}(\text{muonic Li})}{E_{\text{ion}}(\text{normal Li})} = \frac{\mred}{m_e} \approx 207 , \] so the muonic ionization energy is about \(207\) times larger.
In the source manuscript this solution concluded "(c)" while the key said "(a)", with a note that the answer did not match. The physics is unambiguous: heavier orbiting particle \(\Rightarrow\) larger reduced mass \(\Rightarrow\) larger binding, so the factor is \(\approx\!207\) times larger. The option lettering above has been rewritten so that (a) is the correct statement. Cross-check against the official CSIR key before print.
Solution. Energy and momentum conservation for photon capture by a free atom: \[ h\nu + Mc^{2} = \sqrt{p^{2}c^{2} + (M+\Delta M)^{2}c^{4}}, \qquad p = \frac{h\nu}{c}. \] Squaring and cancelling \((h\nu)^{2}\): \[ 2h\nu M c^{2} = \left[(M+\Delta M)^{2} - M^{2}\right]c^{4} = \left(2M\Delta M + \Delta M^{2}\right)c^{4}, \] \[ h\nu = \Delta M c^{2}\left(1 + \frac{\Delta M}{2M}\right) \quad\Longrightarrow\quad \nu = \frac{\Delta M c^{2}}{h}\left(1 + \frac{\Delta M}{2M}\right). \]
Note that the algebra gives \(\left(1+\frac{\Delta M}{2M}\right)\), i.e.\ option (d) as the options are lettered above. The source manuscript listed the answer as (d). Retain (d) and verify the option order against the original paper.
Solution. \(1/\lambda \propto \mred Z^{2}\). Here \(Z=2\) so \(Z^{2}=4\), and the reduced mass changes only in the fourth decimal place, so \[ \lambda' \approx \frac{\lambda}{4}. \]
Solution. For hydrogen \(\mred^{\mathrm H} \approx m_e\) (correction \(\sim 1/1836\)). For muonium (\(\mu^{+}e^{-}\)) with \(m_\mu = 100\,m_e\) from the given numbers, \[ \mred^{\mu} = \frac{m_e m_\mu}{m_e+m_\mu} = \frac{100}{101}m_e = 0.990\,m_e . \] Since \(\nu \propto \mred\), \[ \frac{\nu_\mu - \nu_{\mathrm H}}{\nu_{\mathrm H}} = \frac{\mred^{\mu}}{\mred^{\mathrm H}} - 1 \approx -0.01 . \] The magnitude is \(0.01\) and the sign is negative — muonium levels are slightly less deeply bound. Choose the option matching \(-0.01\).
Solution. The ionization potential of the molecule is the energy of \[ \text{H}_{2} \longrightarrow \text{H}_{2}^{+} + e^{-} . \] Energy is a state function, so route it through atoms — a Born–Haber cycle:
| Step | Process | Energy |
|---|---|---|
| Break the H2 bond | \(\text{H}_{2} \to \text{H} + \text{H}\) | \(+4.478\,\mathrm{eV}\) |
| Ionize one atom | \(\text{H} \to \text{H}^{+} + e^{-}\) | \(+13.6\,\mathrm{eV}\) |
| Form the molecular ion | \(\text{H}^{+} + \text{H} \to \text{H}_{2}^{+}\) | \(-2.651\,\mathrm{eV}\) |
| Total | \(\text{H}_{2} \to \text{H}_{2}^{+} + e^{-}\) | \(\mathbf{+15.427\,\mathrm{eV}}\) |
Hence \(I(\text{H}_{2}) = 15.427\,\mathrm{eV}\).
Written compactly the cycle says \[ I(\text{H}_{2}) = I(\text{H}) + D(\text{H}_{2}) - D(\text{H}_{2}^{+}), \] so the molecular ionization potential exceeds the atomic one precisely because the neutral molecule is more strongly bound than the ion — the electron removed was a bonding electron. The statement that the two bond lengths are equal is what licenses the cycle: it lets you ignore any change in nuclear kinetic energy, so the process is vertical.
Solution. \(\mred = m_e/2\) for positronium and \(E \propto \mred\), so the binding energy is \(13.6\,\mathrm{eV}/2 = 6.8\,\mathrm{eV}\).
Solution. \(E \propto \mred = \dfrac{m_e M}{m_e+M}\), which increases monotonically with \(M\). A lighter proton means a smaller \(\mred\), hence a smaller binding energy — the level lies less deep, so the ground state energy is less negative in magnitude. Explicitly, with \(M = 183.6\,m_e\), \[ \mred = \frac{183.6}{184.6}m_e = 0.9946\,m_e \quad\text{versus}\quad 0.99946\,m_e , \] so \(\lvert E_1\rvert\) falls by about \(0.5\%\).
Solution. The collision can excite the atom only to a level whose excitation energy does not exceed \(13.377\,\mathrm{eV}\): \[ 13.6\left(1-\frac{1}{n^{2}}\right) \le 13.377 \;\Longrightarrow\; \frac{1}{n^{2}} \ge 0.0164 \;\Longrightarrow\; n^{2} \le 61 \;\Longrightarrow\; n = 7 . \] Since \(r \propto n^{2}\), the radius increases by a factor \(n^{2} = 49\).
Solution. \[ R = \Ry\frac{\mred}{m_e} = \frac{\Ry}{1 + m_e/M} \approx \Ry\left(1 - \frac{m_e}{M}\right). \] Comparing with \(R = \Ry(1-2.5\times 10^{-4})\) gives \[ \frac{m_e}{M} = 2.5\times 10^{-4}. \]