Laser

Solution. Optical pumping drives the transition \(1\to2\), but the same radiation also drives stimulated emission \(2\to1\), and \(B_{12}=B_{21}\) makes the two equally probable. As \(N_{2}\) grows the downward rate catches up with the upward rate, and at saturation they are equal — so the populations can at best be equalised, never inverted. This is equation \(\eqref{eq:two-level-result}\) in words.

Solution. In a two-level system population inversion is not achievable at any pump power. Under strong resonant radiation the populations saturate towards equality, so the best that can be reached is \[ N_{g} \approx N_{e} \approx \frac{N}{2}. \]

Solution. A state of lifetime \(\tau\) has an energy uncertainty \(\Delta E \approx \hbar/\tau\) (equivalently \(h/\Delta t\), to within \(2\pi\)): \[ \Delta E = \frac{h}{\Delta t} = \frac{6.625\times10^{-34}}{10^{-9}} = 6.625\times 10^{-25}\,\mathrm{J} = \frac{6.625\times10^{-25}}{1.6\times10^{-19}}\ \eV = 4.14\times10^{-6}\ \eV . \] Using \(\hbar\) instead of \(h\) gives \(6.6\times10^{-7}\,\eV\). Either way the order of magnitude is \(10^{-6}\ \eV\).

Solution. This is a multiple-select question, and both (a) and (b) are required.

(a) The lasing transition in the figure is \(1\to0\), so level \(1\) is the upper laser level and must be metastable — its lifetime has to exceed that of level \(2\), from which it is fed, otherwise atoms leave faster than they arrive and no population accumulates.

(b) Level \(0\) is the lower laser level, so gain on \(1\to0\) requires \(N_{1}>N_{0}\). That is population inversion, without which there is no amplification at all.

Option (c) compares the lifetime of level \(2\) with the ground state and is irrelevant; option (d) demands inversion on the wrong pair of levels.

Solution. Level \(1\) is filled by decay from level \(2\) and emptied by decay to level \(0\): \[ \frac{\dd N_{1}}{\dd t} = \frac{N_{2}}{\tau_{2}} - \frac{N_{1}}{\tau_{1}} = A_{21}N_{2} - A_{10}N_{1} . \] In the steady state \(\dd N_{1}/\dd t = 0\), so \[ N_{1} = \frac{A_{21}}{A_{10}}N_{2} = \frac{2\times10^{7}}{10^{8}}\times10^{14} = 2\times10^{13}\ \mathrm{cm}^{-3} . \] Hence \(x = 2\).

Solution. The lasing atom in a He–Ne laser is neon, so \(A = 20\), not \(4\). Writing \(\nu = c/\lambda\), \[ \Delta\nu = \frac{2\nu}{c}\sqrt{\frac{\alpha}{A}} = \frac{2}{\lambda}\sqrt{\frac{\alpha}{A}} = \frac{2}{633\times10^{-9}}\sqrt{\frac{3.44\times10^{6}}{20}} . \] Now \(\sqrt{3.44\times10^{6}/20} = \sqrt{1.72\times10^{5}} = 414.7\), so \[ \Delta\nu = \frac{2\times414.7}{633}\times10^{9} \approx 1.3\times 10^{9}\,\mathrm{Hz}, \] giving \(n \approx 1.3\).

Solution. Thermal (stimulated) emission dominates when the ratio \(\eqref{eq:sp-st-ratio}\) falls below unity. At \(300\,\mathrm{K}\), \[ \frac{\hbar}{k_{B}T} = \frac{1.054\times10^{-34}}{1.38\times10^{-23}\times300} = 2.55\times 10^{-14}\,\mathrm{s} . \] Test the two plausible options.

For \(\nu = e13\,\mathrm{Hz}\), \(\omega = 6.3\times 10^{13}\,\mathrm{rad}\,\mathrm{s}^{-1}\) and \(\hbar\omega/k_{B}T = 1.6\): \[ \frac{A_{21}}{B_{21}u(\omega)} = e^{1.6}-1 = 4.95-1 \approx 4 > 1, \] so spontaneous emission still wins.

For \(\nu = e11\,\mathrm{Hz}\), \(\omega = 6.3\times 10^{11}\,\mathrm{rad}\,\mathrm{s}^{-1}\) and \(\hbar\omega/k_{B}T = 1.6\times10^{-2}\): \[ \frac{A_{21}}{B_{21}u(\omega)} = e^{0.016}-1 \approx 0.016 \ll 1, \] so thermal stimulated emission dominates. The answer is \(e11\,\mathrm{Hz}\).

Solution. In the \(\rho(\nu)\) convention, \[ \frac{A_{21}}{B_{21}} = \frac{8\pi h\nu^{3}}{c^{3}}, \qquad \tau = \frac{1}{A_{21}} = \frac{c^{3}}{8\pi h\nu^{3}B_{21}} = \frac{\lambda^{3}}{8\pi h B_{21}} . \] With \(\lambda = 3000\times 10^{-10}\,\mathrm{m}\) and \(B_{21} = 2.1\times 10^{19}\,\mathrm{m}^{3}\,\mathrm{W}^{-1}\,\mathrm{s}^{-3}\), \[ \tau = \frac{(3000\times10^{-10})^{3}} {8\pi\times6.6\times10^{-34}\times2.1\times10^{19}} = \frac{2.7\times10^{-20}}{3.5\times10^{-13}} = 7.7\times10^{-8}\ \mathrm{s} \approx 80\,\mathrm{ns}. \]

Solution. With \(n_{0}=1\) and \(\Delta E = 10.2\,\mathrm{eV}\), the \(u(\omega)\) convention gives \[ \frac{B_{21}}{A_{21}} = \frac{\pi^{2}c^{3}}{\hbar\omega^{3}n_{0}^{3}} = \frac{\hbar^{2}\pi^{2}c^{3}}{(\Delta E)^{3}n_{0}^{3}} , \] using \(\omega = \Delta E/\hbar\). Substituting \(\Delta E = 10.2\times1.6\times10^{-19}\ \mathrm{J}\), \[ \frac{B_{21}}{A_{21}} = 0.67\times 10^{12}\,\mathrm{m}^{3}\,\mathrm{J}^{-1}\,\mathrm{s}^{-1} \approx 0.683\times 10^{12}\,\mathrm{m}^{3}\,\mathrm{J}^{-1}\,\mathrm{s}^{-1} . \] The lifetime \(1.6\,\mathrm{ns}\) is not needed — it is a distractor, since \(B/A\) depends only on \(\omega\) and \(n_{0}\).

Solution. The mode separation is \[ \delta\nu = \frac{c}{2dn_{0}} = \frac{3\times10^{8}}{2\times20\times10^{-2}\times1} = 750\,\mathrm{MHz} . \] That alone identifies option (c): no other option has \(\delta\nu = 750\,\mathrm{MHz}\).

For completeness, the mode width follows from the cavity lifetime, \[ t_{c} = \frac{2n_{0}d}{c\,\ln\!\left(1/R_{1}R_{2}e^{-2\alpha d}\right)} = \frac{0.4}{3\times10^{8}\times\ln(1/0.98)} = \frac{0.4}{3\times10^{8}\times0.0202} = 6.6\times 10^{-8}\,\mathrm{s}, \] \[ \Delta\nu_{p} = \frac{1}{2\pi t_{c}} = 2.4\,\mathrm{MHz}. \]

Solution. From \(\nu = mc/2L\), with the mode number \(m\) fixed, \[ \Delta\nu = -\frac{mc}{2L^{2}}\Delta L = -\frac{\nu}{L}\Delta L = -\frac{c}{\lambda L}\Delta L , \] so \[ \Delta L = -\frac{\lambda L}{c}\Delta\nu = -\frac{0.5\times10^{-6}\times1\times100}{3\times10^{8}} = -0.167\times10^{-12}\ \mathrm{m}. \] The negative sign says the cavity length must increase to bring the frequency down — the frequency was found to be too high by \(100\,\mathrm{Hz}\), and \(\Delta\nu\) here is that excess. The magnitude of the retuning is \(0.167\ \mathrm{pm}\): a fraction of an atomic diameter, which is why laser cavities need piezoelectric control.

Solution. Lasing continues until the inversion is exhausted, i.e.\ until the two populations are equal. Starting from \(N_{2}-N_{1} = 3\times10^{20}\), \[ E = \left(\frac{N_{2}-N_{1}}{2}\right)h\nu = \frac{3\times10^{20}}{2}\times2\times1.6\times10^{-19}\ \mathrm{J} = 48\,\mathrm{J}. \]

Solution. The spacing between adjacent levels near \(n\) is \[ \Delta E_{n} = \left|\frac{\dd E_{n}}{\dd n}\right| = \frac{2\times13.6}{n^{3}}\ \eV . \] To excite \(n=100\) selectively, the laser linewidth must be smaller than the gap to the neighbouring level: \[ \Delta\nu = \frac{\Delta E_{n}}{h} = \frac{2\times13.6}{(100)^{3}} \times\frac{1.6\times10^{-19}}{6.626\times10^{-34}} = 6.5\,\mathrm{GHz}. \]

Solution. The frequency separation between adjacent cavity modes is \[ \Delta\nu = \frac{c}{2L} = \frac{3\times10^{8}}{2\times35\times10^{-2}} = 0.42\times10^{9}\ \mathrm{Hz} = 0.42\,\mathrm{GHz}. \] The gain bandwidth is \(1.3\,\mathrm{GHz}\), so the number of modes falling inside it is \[ \frac{1.3}{0.42} = 3.1 \quad\longrightarrow\quad 3 \text{ modes}. \]

Solution. In equilibrium the number of upward transitions equals the number of downward transitions.

Rate of upward transitions \(= P_{0}nW_{0\to1}\) — absorption only, and it requires a photon to be present, hence the factor \(n\).

Rate of downward transitions \(=\) spontaneous \(+\) stimulated \(= P_{1}W_{1\to0} + P_{1}nW_{1\to0}\) — the spontaneous part carries no factor of \(n\), because it happens whether photons are present or not.

Equating, \[ P_{0}nW_{0\to1} = P_{1}W_{1\to0} + P_{1}nW_{1\to0} . \]

Solution. [A] Level \(2\) is filled at the constant pump rate \(R\) and emptied with lifetime \(\tau\): \[ \frac{\dd N_{2}}{\dd t} = R - \frac{N_{2}}{\tau} . \] Integrating with \(N_{2}(0)=0\), \[ N_{2}(t) = R\tau\left(1-e^{-t/\tau}\right), \] which saturates at \(R\tau\) as \(t \to \infty\).

[B] “Transparency” means zero gain, i.e.\ \(N_{2}=N_{1}=N/2\) in this three-level system. The pump must replace atoms decaying out of level \(2\) at the rate \(N_{2}/\tau\), each costing a photon of energy \(hc/\lambda\): \[ P = \frac{N}{2}\frac{h\nu}{\tau} = \frac{N}{2}\frac{hc}{\lambda\tau} = \frac{1.6\times10^{19}}{2} \times\frac{6.6\times10^{-34}\times3\times10^{8}} {0.7\times10^{-6}\times3\times10^{-3}} = 754\,\mathrm{W}\,\mathrm{cm}^{-3} , \] that is \(P = 0.76\,\mathrm{kW}\) per cubic centimetre.

Solution. Level \(2\) is fed by the pump and decays with \(\tau_{21}\); level \(1\) is fed from level \(2\) and decays with \(\tau_{1}\): \[ \frac{\dd N_{2}}{\dd t} = R - \frac{N_{2}}{\tau_{21}}, \qquad \frac{\dd N_{1}}{\dd t} = \frac{N_{2}}{\tau_{21}} - \frac{N_{1}}{\tau_{1}} . \] At equilibrium both derivatives vanish, so \[ N_{2} = R\tau_{21} = 10^{20}\times20\times10^{-9} = 2\times10^{12}\ \mathrm{cm}^{-3}, \] \[ N_{1} = \frac{\tau_{1}N_{2}}{\tau_{21}} = \frac{10^{-6}\times2\times10^{12}}{20\times10^{-9}} = 10^{14}\ \mathrm{cm}^{-3} . \]

Solution. Two beams cross the plane \(P\): the forward wave travelling from \(M_{1}\) to \(M_{2}\), and the backward wave returning. Since \(R_{1}=1\), whatever arrives at \(M_{1}\) is reflected completely, so both waves have intensity \(I_{1}\) at \(x=0\).

The forward wave has been amplified over the distance \(x\) it has already travelled: \(I_{f}(x) = I_{1}e^{\alpha x}\).

The backward wave has yet to travel the distance \(x\) before reaching \(M_{1}\), where its intensity will be \(I_{1}\); so at \(P\) it is smaller by exactly that amplification: \(I_{b}(x) = I_{1}e^{-\alpha x}\).

Energy density is intensity divided by the speed of propagation, and the two contributions add: \[ u = \frac{I_{f}+I_{b}}{c} = \frac{I_{1}}{c}\left(e^{\alpha x}+e^{-\alpha x}\right). \]

Solution. The net rate of stimulated emission is governed by the population difference, since every stimulated emission from the upper level competes with an absorption from the lower one: \[ \frac{\dd N_{2}}{\dd t} = B_{21}\left(N_{2}-N_{1}\right)u(\nu). \] Note that \(\mathrm{m}^{3}\,\mathrm{W}^{-1}\,\mathrm{s}^{-3} \equiv \mathrm{m}^{3}\,\mathrm{J}^{-1}\,\mathrm{s}^{-2}\) and \(\mathrm{J}\,\mathrm{m}^{-3}\,\mathrm{Hz}^{-1} \equiv \mathrm{J}\,\mathrm{s}\,\mathrm{m}^{-3}\), so the product \(B_{21}u\) has units of \(\mathrm{s}^{-1}\) as it must. With \(N_{2}-N_{1} = (3-0.7)\times10^{18} = 2.3\times10^{18}\), \[ \frac{\dd N_{2}}{\dd t} = 2.3\times10^{18}\times3.0\times10^{5}\times9.0 = 6.2\times10^{24}\ \mathrm{s}^{-1} . \] The mantissa \(6.2\) matches option (a).

Solution. For \(3\to4\) to lase, level \(3\) must be metastable and level \(4\) must empty quickly. Reading the requirements off the diagram:

• \(\Gamma_{34}\) must be the smallest, so that population accumulates in level \(3\).

• \(\Gamma_{41}\) must be large, so that level \(4\) — the lower laser level — is drained and stays empty.

• \(\Gamma_{23}\) must exceed \(\Gamma_{21}\), so that the pumped population in level \(2\) is delivered to level \(3\) rather than falling straight back to level \(1\).

Only \(\Gamma_{41} > \Gamma_{23} > \Gamma_{21} > \Gamma_{34}\) satisfies all three.

Solution. The two frequencies produce two interference patterns, slightly displaced from each other. The fringes wash out when the maxima of one fall on the minima of the other, that is when \[ d\sin\theta = n\lambda_{1} \quad\text{coincides with}\quad d\sin\theta = \left(n+\tfrac12\right)\lambda_{2}. \] The relevant length is the beat wavelength of the two lines, \[ \lambda = \frac{\lambda_{1}\lambda_{2}}{\lambda_{1}-\lambda_{2}} = \frac{1}{\dfrac{1}{\lambda_{2}}-\dfrac{1}{\lambda_{1}}} = \frac{c}{\nu_{2}-\nu_{1}} , \] and the pattern first disappears at half of it: \[ \frac{\lambda}{2} = \frac{c}{2(\nu_{2}-\nu_{1})} . \]

Solution. The Doppler width is \[ \frac{\Delta\nu_{0}}{\nu_{0}} = 1.67\sqrt{\frac{2k_{B}T}{mc^{2}}} . \] Both the mean kinetic energy and the rest mass are given in electronvolts, so they may be divided directly: \[ \frac{\Delta\nu_{0}}{\nu_{0}} = 1.67\sqrt{\frac{0.33}{22.5\times10^{9}}} = 1.67\times3.83\times10^{-6} = 6.4\times10^{-6} \approx 6\times10^{-6} . \]

Solution. Level \(E_{2}\) can decay by two routes, to \(E_{0}\) and to \(E_{1}\), and the rates add: \[ \left(A_{20}+A_{21}\right)N_{2} = A_{2}N_{2}, \] \[ A_{2} = A_{20}+A_{21} = \left(1.2\times10^{8} + 0.8\times10^{8}\right)\mathrm{s}^{-1} = 2.0\times 10^{8}\,\mathrm{s}^{-1} . \] The mean radiative lifetime is therefore \[ \tau_{2} = \frac{1}{A_{2}} = \frac{1}{2.0\times10^{8}} = 0.5\times10^{-8}\ \mathrm{s} . \] \(A_{10}\) governs level \(1\) and plays no part.

Solution. Each absorbed photon delivers momentum \(p_{\gamma} = h/\lambda\) along the beam, opposing the atom's motion. The photon subsequently re-emitted by spontaneous emission goes in a random direction, so averaged over many cycles its recoil contributes nothing. The net momentum removed per cycle is therefore \(h/\lambda\). \[ p_{\gamma} = \frac{h}{\lambda} = \frac{6.626\times10^{-34}}{776.5\times10^{-9}} = 8.53\times 10^{-28}\,\mathrm{kg}\,\mathrm{m}\,\mathrm{s}^{-1}, \] \[ m_{K} = 39\times1.66\times10^{-27} = 6.48\times 10^{-26}\,\mathrm{kg}. \] The number of cycles needed to remove the whole momentum \(m_{K}v_{0}\) is \[ N = \frac{m_{K}v_{0}}{p_{\gamma}} = \frac{6.48\times10^{-26}\times130}{8.53\times10^{-28}} = 9880 \approx \boxed{10^{4}} . \]

Solution. A wavelength \(\lambda\) resonates in a cavity of length \(L\) when \(L = n\lambda/2\) for integer \(n\) — that is, when \(L\) is an integer multiple of \(\lambda/2\). For all three wavelengths to oscillate simultaneously, \(L\) must be a common multiple of \[ \frac{\lambda}{2} \in \{225,\ 300,\ 375\}\ \mathrm{nm}. \] Factorising, \[ 225 = 3^{2}\times5^{2}, \qquad 300 = 2^{2}\times3\times5^{2}, \qquad 375 = 3\times5^{3}, \] \[ \mathrm{LCM} = 2^{2}\times3^{2}\times5^{3} = 4500\,\mathrm{nm} = 4.5\,\mathrm{\mu m}. \] So \(L\) must be a multiple of \(4.5\,\mathrm{\mu m}\). Testing the four options: \[ \frac{450}{4.5} = 100\ \checkmark, \qquad \frac{600}{4.5} = 133.3\ \times, \qquad \frac{750}{4.5} = 166.7\ \times, \qquad \frac{1500}{4.5} = 333.3\ \times . \] Only \(L = 450\,\mathrm{\mu m}\) supports all three modes.

Solution. The photon must supply both the excitation energy and the recoil kinetic energy of the atom. Conservation of energy and of momentum give \[ \hbar\omega = \hbar\Delta + \frac{p^{2}}{2m}, \qquad p = \frac{\hbar\omega}{c}, \] so \[ \hbar\omega = \hbar\Delta + \frac{\hbar^{2}\omega^{2}}{2mc^{2}} . \] The recoil term is minute, so solve by one step of iteration: to zeroth order \(\omega \approx \Delta\), and substituting that inside the small correction only, \[ \omega = \Delta + \frac{\hbar\Delta^{2}}{2mc^{2}} = \boxed{\Delta\left(1 + \frac{\hbar\Delta}{2mc^{2}}\right)} . \]