Corrected exercises: modern physics (203-SN3-RE)

Sun: $\lambda_{max}=\frac{2.898\times 10^{-3}}{5800}\approx 5.00\times 10^{-7}$ m, that is 500 nm, right in the middle of the visible range, in the green. Human body: $\lambda_{max}=\frac{2.898\times 10^{-3}}{310}\approx 9.35\times 10^{-6}$ m, that is 9.3 μm, in the long-wave infrared, completely invisible to the eye but perfectly picked up by a thermal camera. The coincidence between the solar maximum and the peak sensitivity of the eye, around 555 nm, is of course no coincidence: the eye evolved under this light and adjusted to what was most abundant.

$\lambda_{max}=\frac{2.898\times 10^{-3}}{2500}\approx 1.16\times 10^{-6}$ m, that is 1.16 μm: the maximum falls in the INFRARED, outside the visible range. An incandescent bulb therefore emits most of its radiation as invisible heat, and only the short-wavelength tail of its distribution reaches the visible; its luminous efficiency tops out around 5%. To do better, the temperature would have to go up, but tungsten melts at 3695 K: blackbody physics doomed the filament bulb long before regulations got involved.

$P=\sigma A(T^{4}-T_{0}^{4})=(5.67\times 10^{-8})(1.8)(306^{4}-293^{4})\approx 1.4\times 10^{2}$ W. About 140 W, more than the basal metabolic rate of an adult at rest, which explains why we cool down so quickly when unclothed, even in a room at 20 °C. The net power is proportional to the DIFFERENCE of the fourth powers, never to the difference of the temperatures: this is why a gap of only 13 degrees is enough to carry away that much energy. Clothing does not work by suppressing this radiation but by placing an intermediate surface in the way, colder on its outer side.

Classical physics treated the radiation as a set of oscillation modes in a cavity, and the equipartition theorem gave each one the same average energy $kT$. But the number of modes grows without limit as the wavelength decreases: the total diverged, and one predicted that an open oven should emit a deadly burst of ultraviolet. Planck assumed that the energy of a mode of frequency $f$ can only be $nhf$, an integer multiple. At high frequencies, the FIRST quantum $hf$ is already much larger than the typical thermal energy $kT$: these modes simply cannot be excited, they stay frozen and stop contributing. The divergence disappears, and the calculated curve matches experiment exactly. Planck regarded this as a mathematical device; it was Einstein who, five years later, made it a physical reality with the photon.

The wavelength of maximum emission is inversely proportional to $T$: it is therefore halved, to 250 nm, in the ultraviolet, and the star looks blue-white since only the edge of its distribution is left to us. The power per square metre goes as $T^{4}$: it is multiplied by $2^{4}=16$. The same law ranks the colours of heated objects: a red ember radiates around 1000 K, white-hot iron exceeds 1500 K, and its pale colour signals not a decorative nuance but a maximum emission shifted toward the blue and a power five times greater. In astronomy, the colour of a star is a thermometer.

$E=\frac{hc}{\lambda}=\frac{(6.63\times 10^{-34})(3.0\times 10^{8})}{500\times 10^{-9}}\approx 4.0\times 10^{-19}$ J, that is $\frac{3.98\times 10^{-19}}{1.6\times 10^{-19}}\approx 2.5$ eV. (Shortcut: $\frac{1240}{500}=2.48$ eV.) Visible photons all lie between 1.8 and 3.1 eV: the same scale as chemical bonds, which is why visible light drives photosynthesis and vision.

$E=hf=(6.63\times 10^{-34})(1.0\times 10^{8})=6.6\times 10^{-26}$ J $\approx 4.1\times 10^{-7}$ eV: six million times less than a visible photon. A radio receiver absorbs streams of billions of billions of photons: the grain is so fine that the energy looks perfectly continuous, just as a beach seen from afar looks smooth.

Energy of a red photon: $E=\frac{hc}{\lambda}=\frac{1.989\times 10^{-25}}{650\times 10^{-9}}\approx 3.1\times 10^{-19}$ J. Rate: $N=\frac{P}{E}=\frac{1.0\times 10^{-3}}{3.06\times 10^{-19}}\approx 3.3\times 10^{15}$ photons per second: three million billion, for the most modest of laser pointers.

Everything depends on the ratio between the quantum $hf$ and the energies involved. In radio, $hf\approx 10^{-7}$ eV is tiny compared with everything: exchanges of billions of photons average out into a continuous wave, and the classical description is enough. An X-ray photon carries $10^{4}$ to $10^{5}$ eV: ONE photon is enough to ionize an atom, break a bond, or darken a grain of film: the interaction is an individual, discrete event that no continuous wave can describe. The classical-quantum boundary is not a question of size, but of relative graininess.

First, the existence of a frequency THRESHOLD: below $f_{0}$, no electron comes out, even under blinding light, whereas a wave should be able to accumulate the necessary energy by persisting. Second, the INSTANTANEOUS emission, even in very faint light, where a diluted wave should take minutes to charge up an electron. With the photon, everything becomes clear: the exchange takes place in packets $hf$, and an electron absorbs ONE photon; if $hf<W_{0}$, the packet is too small and no accumulation is possible; if it is enough, ejection is immediate. The energy of light is not in its intensity but in its frequency: this is Einstein's reversal (1905).

Photon of 400 nm: $E=\frac{1240}{400}=3.1$ eV. Energy balance: $E_{k,max}=E-W_{0}=3.1-2.0=1.1$ eV. Stopping potential: $V_{s}=\frac{E_{k,max}}{e}=1.1$ V: the voltage that cancels the current measures, directly in volts, the energy in eV.

$\lambda_{0}=\frac{hc}{W_{0}}=\frac{1240}{2.0}=620$ nm: an orange-red. At 700 nm, each photon carries only $\frac{1240}{700}\approx 1.8$ eV $<2.0$ eV: NO electron comes out, whatever the intensity; multiplying photons that are too weak does not manufacture a photon that is strong enough.

Doubling the intensity at 400 nm doubles the NUMBER of photons per second: twice as many electrons are ejected (double the current), but each with the same maximum energy of 1.1 eV: the stopping potential does not move. Switching to 350 nm changes the photon itself: $E=\frac{1240}{350}\approx 3.5$ eV, so $E_{k,max}\approx 1.5$ eV and $V_{s}\approx 1.5$ V, with a comparable current. Intensity and frequency act on two different knobs: this is exactly the particle signature.

$E_{1}=-13.6$ eV, $E_{2}=-\frac{13.6}{4}=-3.40$ eV, $E_{3}=-\frac{13.6}{9}\approx-1.51$ eV. The zero is the reference: electron torn free, at rest, infinitely far from the proton. Bound energies are negative because energy MUST be supplied to free the electron: the minus sign measures the depth of the well.

$\Delta E=E_{3}-E_{2}=-1.51-(-3.40)=1.89$ eV. Wavelength: $\lambda=\frac{1240}{1.89}\approx 656$ nm: the RED line of the Balmer series (H$\alpha$), the one that colours nebulae pink. A mental calculation recovers the colour of deep space: this is the triumph of the Bohr model (1913).

From the ground state, $13.6$ eV must be supplied. Maximum wavelength: $\lambda=\frac{1240}{13.6}\approx 91$ nm: far ultraviolet. Any shorter photon also ionizes (the surplus goes into the kinetic energy of the electron); any longer photon cannot, unless it finds the atom already excited.

The electron does not have access to a continuum of energies: only the levels $E_{n}$ exist, so only the DIFFERENCES $E_{n_{i}}-E_{n_{f}}$ can come out as photons: a comb of lines, the fingerprint of the element. In absorption, the mechanism works in reverse: the atom can swallow ONLY the photons whose energy matches an allowed jump exactly: same differences, same wavelengths, dark lines on a continuous background. This is how we read the composition of the Sun without going there: its absorption lines are the same as those of our hydrogen lamps.

$\frac{1}{2}mv^{2}=eV$ gives $v=\sqrt{\frac{2(1.6\times 10^{-19})(100)}{9.11\times 10^{-31}}}\approx 5.9\times 10^{6}$ m/s. Momentum: $p=mv\approx(9.11\times 10^{-31})(5.93\times 10^{6})\approx 5.4\times 10^{-24}$ kg·m/s. Hence $\lambda=\frac{h}{p}=\frac{6.63\times 10^{-34}}{5.4\times 10^{-24}}\approx 1.2\times 10^{-10}$ m: 0.12 nm.

$p=mv=(0.145)(40)=5.8$ kg·m/s, so $\lambda=\frac{6.63\times 10^{-34}}{5.8}\approx 1.1\times 10^{-34}$ m.

The wavelength of the electron (100 V, barely a battery) falls RIGHT at the atomic scale: the planes of a crystal, spaced about $10^{-10}$ m apart, form a natural diffraction grating for it: this is the Davisson-Germer experiment (1927), which confirmed de Broglie by observing electrons interfering like X-rays. The ball, for its part, is at $10^{-34}$ m: twenty-four orders of magnitude below the size of an atom; no opening in the universe is fine enough to reveal its wave. The duality is universal in principle, invisible in practice as soon as $h$ is crushed by $p$.

Diffraction forbids resolving details smaller than about one wavelength: visible optics runs into the wall near 400 nm, a few thousand atoms. The electrons of part a), with 0.12 nm under a modest voltage (and far less still under 100 kV), lower this limit by several orders of magnitude: the electron microscope does not win through better lenses, it wins by changing the wave. To see small, first illuminate short.

$\frac{h}{m_{e}c}=\frac{6.63\times 10^{-34}}{(9.11\times 10^{-31})(3.0\times 10^{8})}\approx 2.43\times 10^{-12}$ m, that is 2.43 pm. At 90°, $\cos\theta=0$ and $\Delta\lambda=2.43$ pm. At 180°, backscattering, $\cos\theta=-1$ and $\Delta\lambda=4.85$ pm, the maximum possible. This is the signature of a collision: the photon gives up the most energy when it bounces straight back, exactly like a ball rebounding head-on.

Incident photon: $E=\frac{hc}{\lambda}=\frac{(6.63\times 10^{-34})(3.0\times 10^{8})}{71.0\times 10^{-12}}\approx 2.80\times 10^{-15}$ J, that is $\frac{2.80\times 10^{-15}}{1.6\times 10^{-19}}\approx 17.5$ keV. Scattered photon, with wavelength $71.0+2.43=73.4$ pm: $E'\approx 2.71\times 10^{-15}$ J, that is 16.9 keV. By conservation of energy, the electron carries off the difference: $E-E'\approx 5.8\times 10^{2}$ eV, about 578 eV. Note how modest the transfer is, barely 3%: the photon bounces off a particle far more massive than it is, at least in terms of equivalent inertia.

The shift is ABSOLUTE, always a few picometres, never proportional to $\lambda$. For an X-ray of 71 pm, it represents $\frac{2.43}{71.0}\approx 3.4$%, easily measurable. For visible light at 500 nm, it represents only $\frac{2.43\times 10^{-12}}{5.0\times 10^{-7}}\approx 4.9\times 10^{-6}$, five millionths, undetectable with the means of the time and with an ordinary spectrometer today. This is why the effect stayed invisible as long as people worked only with visible light, and why X-rays were needed to reveal it: one needs a photon whose energy is comparable to the mass energy of the electron.

The photoelectric effect established that light exchanges its energy in packets $hf$. It remained possible to save the idea of a wave by attributing the graininess to the metal rather than to the light. Compton closes that door: to account for the angular shift, one must apply conservation of MOMENTUM to the photon, with $p=\frac{h}{\lambda}$, exactly as in a two-body elastic collision in mechanics. The photon is therefore not just a grain of energy, it is a grain with momentum, which bounces, transfers and recoils. This is the point where the photon stops being a calculating convenience and becomes a particle, and Compton receives the Nobel Prize in 1927.

In the atom: $p\approx\frac{\hbar}{\Delta x}=\frac{1.055\times 10^{-34}}{1.0\times 10^{-10}}\approx 1.06\times 10^{-24}$ kg·m/s, hence $E_{k}=\frac{p^{2}}{2m_{e}}\approx 6.1\times 10^{-19}$ J, about 3.8 eV. This is the same order of magnitude as the 13.6 eV that actually bind the electron: the uncertainty principle therefore explains the SIZE of the atom. Squeezing the electron closer to the nucleus would lower its potential energy, but at the price of a confinement kinetic energy that grows as $\frac{1}{\Delta x^{2}}$, hence faster: there is an equilibrium radius, and the atom does not collapse. In the nucleus: $p\approx\frac{1.055\times 10^{-34}}{1.0\times 10^{-15}}\approx 1.06\times 10^{-19}$ kg·m/s, and since this value is ultrarelativistic, $E\approx pc\approx 3.2\times 10^{-11}$ J, nearly 200 MeV. Yet no nuclear force binds anything with such an energy, the binding energies per nucleon being of the order of 8 MeV. Conclusion: an electron can NOT reside in a nucleus, which definitively ruled out the model of nuclear electrons and, after the discovery of the neutron in 1932, imposed the nucleus made of protons and neutrons. The electrons of beta radioactivity do not pre-exist in the nucleus, they are created at the moment of the decay.

Carbon-14 decays continuously, but living things renew it by breathing and eating: the balance between losses and supplies fixes the proportion. At death, exchanges stop while the losses continue: the proportion decays exponentially, and its present level measures the elapsed time. A dead organism is an hourglass that is never turned over again.

$25\ \%=\frac{1}{4}=\left(\frac{1}{2}\right)^{2}$: exactly TWO half-lives. Age: $2\times 5730=11{,}460$ years: the end of the last glaciation. Each half-life halves what remains, whatever the starting point.

$\lambda=\frac{\ln 2}{T}=\frac{0.693}{5730}\approx 1.21\times 10^{-4}$ yr$^{-1}$. Then $N=N_{0}e^{-\lambda t}$ gives $t=\frac{\ln(N_{0}/N)}{\lambda}=\frac{\ln(1/0.30)}{1.21\times 10^{-4}}=\frac{1.204}{1.21\times 10^{-4}}\approx 9950$ years. Consistent: 30% lies between one half-life (50%) and two (25%), hence between 5730 and 11,460 years, closer to the second.

At 66 million years, more than 11,000 half-lives have elapsed: the remaining fraction is $\left(\frac{1}{2}\right)^{11{,}500}$, zero to any physical precision: literally not one atom is left to count, and the slightest gram of modern contamination dominates the signal. At 50 years, conversely, only about 0.6% of the stock has decayed: the change is drowned in the uncertainties. The method is reliable, roughly, from a few centuries to 50,000 years: the clock must have run, but not run out.

One needs an isotope whose half-life is comparable to the age being sought: for dinosaurs, geological pairs such as potassium-argon ($T\approx 1.25$ billion years) or uranium-lead, applied to the rocks that bracket the fossil. General rule: choose your hourglass to match the duration to be measured, just as you do not time a marathon with a three-minute egg timer.

Sum of the free nucleons: $2(1.00728)+2(1.00866)=4.03188$ u. Mass defect: $\Delta m=4.03188-4.00151=0.03037$ u. Binding energy: $E=(0.03037)(931.5)\approx 28.3$ MeV, that is $\frac{28.3}{4}\approx 7.07$ MeV per nucleon. In other words, assembling a helium nucleus releases 28 MeV, and the same amount would have to be supplied to take it apart. Relative to the total, the defect represents only 0.75% of the mass: the mass of an object is not the sum of the masses of its parts, it includes the binding energy, counted negatively.

A nucleus to the LEFT of the maximum can gain binding energy per nucleon by fusing with another: it moves toward the peak, and the excess is released. A nucleus to the RIGHT gains instead by splitting in two pieces closer to iron: this is fission. The two processes therefore go in the same direction, that of the climb toward the maximum, but start from opposite sides. Iron-56, at the top, is the most tightly bound nucleus there is: neither fusing it nor splitting it yields anything, one must on the contrary pay in both cases. This is why a massive star, which makes heavier and heavier elements through successive fusions, stops dead when its core becomes iron: the energy source that held it up against its own gravity goes out, and the collapse that follows produces a supernova. All the elements heavier than iron were, moreover, forged in these cataclysms, against the energy current.

$\Delta m=\frac{200}{931.5}\approx 0.215$ u out of the 235 at the start, about 0.09% of the mass. Number of nuclei in one kilogram: $\frac{1000}{235}\times 6.022\times 10^{23}\approx 2.56\times 10^{24}$. Total energy: $(2.56\times 10^{24})(200\times 10^{6})(1.6\times 10^{-19})\approx 8.2\times 10^{13}$ J/kg. Coal provides $3\times 10^{7}$ J/kg: the ratio is of the order of $2.7\times 10^{6}$, almost three million. One kilogram of uranium replaces about 2700 tonnes of coal, and this disproportion rests entirely on the fact that nuclear energies are counted in MeV whereas chemical energies are counted in eV.

$\frac{1000}{5}\times 6.022\times 10^{23}\approx 1.20\times 10^{26}$ reactions per kilogram, each at 17.6 MeV: $(1.20\times 10^{26})(17.6\times 10^{6})(1.6\times 10^{-19})\approx 3.4\times 10^{14}$ J/kg, about FOUR times the fission. The reason can be read directly on the curve of part b): fission gains $\frac{200}{236}\approx 0.85$ MeV per nucleon, D-T fusion gains $\frac{17.6}{5}\approx 3.5$ MeV, four times more, because the slope of the curve is much steeper on the side of the light nuclei.

$U=\frac{ke^{2}}{r}=\frac{(9.0\times 10^{9})(1.6\times 10^{-19})^{2}}{3.0\times 10^{-15}}\approx 7.7\times 10^{-14}$ J, about 480 keV. Thermal energy at the core of the Sun: $kT=(1.38\times 10^{-23})(1.5\times 10^{7})\approx 2.1\times 10^{-16}$ J, only 1.3 keV: the barrier is nearly FOUR HUNDRED times too high. Classically, the Sun should not shine. Two quantum and statistical effects save the day. First, the Maxwell-Boltzmann distribution has a tail: a tiny but non-zero fraction of the nuclei moves much faster than the average. Next, and above all, TUNNELLING lets a nucleus cross the barrier without having the energy to climb over it, with a minuscule probability which, multiplied by the colossal number of nuclei and by the billions of years available, is more than enough. This improbability is, incidentally, humanity's good fortune: if fusion were easy, the Sun would have burned all its hydrogen in a few million years instead of ten billion.

Mass and energy are two faces of the same quantity: all energy has a mass, all mass is condensed energy, at the exchange rate $c^{2}=9.0\times 10^{16}$ J/kg. This colossal factor explains the asymmetry of everyday life: enormous energies correspond to tiny masses, which is why nobody had noticed the equivalence before 1905.

$\frac{\Delta m}{\Delta t}=\frac{P}{c^{2}}=\frac{3.8\times 10^{26}}{9.0\times 10^{16}}\approx 4.2\times 10^{9}$ kg/s: more than four million tonnes per second, converted into light by the fusion of hydrogen into helium.

Duration: $4.5\times 10^{9}\times 3.16\times 10^{7}\approx 1.4\times 10^{17}$ s. Mass radiated: $(4.22\times 10^{9})(1.42\times 10^{17})\approx 6.0\times 10^{26}$ kg. Fraction: $\frac{6.0\times 10^{26}}{2.0\times 10^{30}}=3.0\times 10^{-4}$: three hundredths of a percent in four and a half billion years. The most extreme regime in the universe costs the Sun almost nothing: there, in a single number, is the power of the exchange rate $c^{2}$.

$\frac{200\times 10^{6}\ \mathrm{eV}}{4\ \mathrm{eV}}\approx 5\times 10^{7}$: nuclear power releases tens of millions of times more energy per reaction than chemistry. The reason lies in the scales of the chapter: chemistry rearranges the ELECTRONS, bound by a few eV; nuclear power rearranges the NUCLEUS, whose bonds are counted in MeV, and the corresponding mass difference, about one thousandth of the mass of the nucleus, leaves as energy through $E=mc^{2}$. One gram of nuclear fuel is thus worth tonnes of chemical fuel: same equation, upper storey of the atomic edifice.

$\gamma(0.60c)=\frac{1}{\sqrt{1-0.36}}=\frac{1}{0.80}=1.25$; $\gamma(0.90c)=\frac{1}{\sqrt{0.19}}\approx 2.29$; $\gamma(0.99c)=\frac{1}{\sqrt{0.0199}}\approx 7.09$. The growth is not steady: $\gamma$ stays close to 1 over almost the whole range, then explodes in the last few per cent. At 60% of the speed of light, the effects do not exceed 25%, which explains why they stayed invisible through three centuries of mechanics.

$\gamma=\frac{1}{\sqrt{1-0.998^{2}}}\approx 15.8$. Without relativity, the muon would travel $d=vt=(0.998)(3.0\times 10^{8})(2.2\times 10^{-6})\approx 659$ m: it would cover only one fifteenth of the trip and practically no muon would reach the ground. With time dilation, its lifetime measured FROM THE GROUND is $\gamma(2.2\ \mu\text{s})\approx 34.8$ μs, and the distance travelled is $\approx 1.04\times 10^{4}$ m, more than 10 km: the muons arrive in abundance, which is indeed measured at sea level. In the frame of the MUON, nothing is dilated, its life does last 2.2 μs; but it is the atmosphere that rushes past, and its thickness is CONTRACTED: $\frac{10{,}000}{15.8}\approx 632$ m, which it crosses in $\frac{632}{0.998\times 3.0\times 10^{8}}\approx 2.1$ μs, roughly one lifetime. The two descriptions, one in dilated time, the other in contracted length, are incompatible in appearance and yet give exactly the same observable prediction. This is the heart of relativity: what is conserved is neither the duration nor the length, but the agreement on the FACTS.

$E=\gamma mc^{2}=(7.089)(0.511)\approx 3.62$ MeV. Kinetic energy: $K=(\gamma-1)mc^{2}=(6.089)(0.511)\approx 3.11$ MeV. The classical formula would give $\frac{1}{2}mv^{2}=\frac{1}{2}(0.511)(0.99)^{2}\approx 0.250$ MeV, twelve times less. At this speed, Newtonian mechanics is not off by a little, it is off by an order of magnitude, which is why every particle accelerator is designed from end to end on the relativistic formulas.

$E=\gamma mc^{2}$ and $\gamma\to\infty$ as $v\to c$: reaching exactly $c$ would require an INFINITE energy for any non-zero mass. No matter how much power is invested, one only nibbles at the decimals: at the Large Hadron Collider, protons reach $0.999999991c$, with $\gamma\approx 7500$, and most of the energy supplied no longer serves to accelerate them but to increase their inertia. The speed of light is not a speed limit in the sense of a rule that could be broken: it is an asymptote built into the structure of spacetime.

At rest, $p=0$ and the relation gives $E^{2}=(mc^{2})^{2}$, hence $E=mc^{2}$: the mass energy is recovered as a special case. For a photon, $m=0$, which leaves $E=pc$, that is $p=\frac{E}{c}=\frac{hf}{c}=\frac{h}{\lambda}$: the momentum of the photon, postulated by de Broglie and verified by Compton, comes straight out of relativity, with no additional assumption. The loop is closed: it is the same relation $p=\frac{h}{\lambda}$ that describes matter as a wave in de Broglie and light as a particle in Compton. It also explains how the photon can carry momentum without having any mass at all, which makes radiation pressure, solar sails and laser cooling of atoms possible.