Corrected exercises: quantum physics, photons, line spectra and duality (203-SN3-RE)

$p=\dfrac{h}{\lambda}=\dfrac{6.63\times 10^{-34}}{550\times 10^{-9}}=1.21\times 10^{-27}$ kg·m/s.

Power received: $P=IA=1361\times 100=1.36\times 10^{5}$ W. Energy of one photon: $E=\dfrac{hc}{\lambda}=3.62\times 10^{-19}$ J. Number per second: $N=\dfrac{P}{E}=3.76\times 10^{23}$ photons/s.

The momentum goes from $+p$ to $-p$: its change is $-2p$, and the sail receives $+2p$. An absorbing sail would receive only $p$: reflecting doubles the thrust, exactly as a bouncing ball transfers more impulse than a ball that sticks.

$F=N\times 2p=\dfrac{IA}{E}\times\dfrac{2E}{c}=\dfrac{2IA}{c}$, since $p=\dfrac{E}{c}$ for a photon. The wavelength drops out of the result. $F=\dfrac{2\times 1361\times 100}{3.00\times 10^{8}}=9.07\times 10^{-4}$ N, the weight of a tenth of a gram.

$a=\dfrac{F}{m}=9.07\times 10^{-5}$ m/s², and over $86\,400$ s: $v=at=7.84$ m/s. That is negligible after one day, but the thrust never stops and uses no fuel: after one year, $2.9$ km/s.

$\dfrac{1}{\lambda}=1.097\times 10^{7}\left(\dfrac{1}{4}-\dfrac{1}{9}\right)=1.524\times 10^{6}$ m⁻¹, so $\lambda=656$ nm: the red line H$\alpha$.

Try successive values: $n_{2}=4$ gives $486$ nm, $n_{2}=5$ gives $434$ nm, $n_{2}=6$ gives $\dfrac{1}{1.097\times 10^{7}(1/4-1/36)}=410$ nm. It is the transition $6\rightarrow 2$.

When $n_{2}\to\infty$, $\dfrac{1}{n_{2}^{2}}\to 0$: $\lambda_{lim}=\dfrac{4}{R_{H}}=3.65\times 10^{-7}$ m $=365$ nm, in the near ultraviolet. No Balmer line has a shorter wavelength.

$\dfrac{1}{\lambda}=1.097\times 10^{7}\times\dfrac{3}{4}$, so $\lambda=122$ nm, in the far ULTRAVIOLET. The whole Lyman series is invisible and absorbed by the atmosphere: it is observed from space.

The energy levels, which go as $-\dfrac{1}{n^{2}}$, get closer together as $n$ increases. The gaps from level $2$ therefore tend toward a limiting value: the emitted photons have energies that get closer and closer, and the lines pile up near the limit of $365$ nm.

$r_{3}=9\times 0.0529=0.476$ nm.

$v_{3}=\dfrac{2.19\times 10^{6}}{3}=7.30\times 10^{5}$ m/s.

$T=\dfrac{2\pi a_{0}}{v_{1}}=\dfrac{2\pi\times 5.29\times 10^{-11}}{2.19\times 10^{6}}=1.52\times 10^{-16}$ s. That is a frequency of $6.6\times 10^{15}$ Hz, of the same order as that of ultraviolet light: this is what made the model plausible.

$\lambda=\dfrac{h}{m_{e}v_{1}}=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 2.19\times 10^{6}}=3.32\times 10^{-10}$ m, and $2\pi a_{0}=3.32\times 10^{-10}$ m. They are equal: the ground-state orbit contains exactly ONE wavelength. In general $2\pi r_{n}=n\lambda_{n}$ is the Bohr condition rewritten, since $m_{e}vr=n\hbar$ is equivalent to $2\pi r=n\dfrac{h}{m_{e}v}$: the electron is a standing wave that closes on itself, like a vibrating string bent into a ring.

$r_{100}=10^{4}\times 0.0529$ nm $=529$ nm $=0.529$ µm, ten thousand times the size of an ordinary atom, almost the size of a bacterium. These giant atoms, very sensitive to electric fields, are used today in quantum optics.

$E_{ion}=13.6\times 2^{2}=54.4$ eV. The electron is attracted by a double charge, at half the distance: it is bound four times as strongly.

$E_{ion}=13.6\times 9=122.4$ eV.

$\Delta E=54.4\left(\dfrac{1}{9}-\dfrac{1}{16}\right)=54.4\times\dfrac{7}{144}=2.64$ eV.

$\lambda=\dfrac{1240}{2.644}=469$ nm, in the BLUE. This line of ionized helium, observed in the Sun and in hot stars, was at first wrongly attributed to hydrogen. Note: $4\rightarrow 3$ belongs to no visible series of hydrogen; it is the factor $Z^{2}=4$ that brings the line back into the visible.

$r_{1}=\dfrac{a_{0}}{Z}$: three times smaller. The energy, for its part, is multiplied by $Z^{2}=9$: the two effects, a stronger charge and a shorter distance, combine.

The kinetic energy gained is $eV=\dfrac{p^{2}}{2m_{e}}$, so $p=\sqrt{2m_{e}eV}$ and $\lambda=\dfrac{h}{p}$. Numerically: $p=\sqrt{2\times 9.11\times 10^{-31}\times 1.60\times 10^{-19}\times 54}=3.97\times 10^{-24}$ kg·m/s and $\lambda=1.67\times 10^{-10}$ m $=0.167$ nm.

$\sin\theta=\dfrac{0.167}{0.215}=0.777$, so $\theta=51°$. Davisson and Germer measured $50°$: the agreement established the wave nature of the electron.

$\lambda\propto\dfrac{1}{\sqrt{V}}$: to divide $\lambda$ by $2$, $V$ must be multiplied by $4$, that is $216$ V. The classic mistake is to double the voltage.

$\lambda=\dfrac{h}{\sqrt{2m_{n}E}}=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.675\times 10^{-27}\times 0.025\times 1.60\times 10^{-19}}}=1.81\times 10^{-10}$ m $=0.181$ nm.

Diffraction is pronounced only if the wavelength is of the same order as the spacing of the obstacles. The distances between atoms are a few tenths of a nanometre: that is precisely the wavelength of these electrons and neutrons, whereas visible light, from $400$ to $700$ nm, is a thousand times too large.

At best, an electron can give up all of its kinetic energy $eV=30.0$ keV to a single photon: $\lambda_{min}=\dfrac{hc}{eV}=\dfrac{1240}{30\,000}=0.0413$ nm.

Because energy is exchanged in photons: a shorter photon would have more energy than the electron that produces it, which conservation of energy forbids. Classical wave theory predicted no sharp cutoff; its existence is direct proof of quantization.

$E=10.2\times 28^{2}=8.00\times 10^{3}$ eV $=8.00$ keV. It is the factor $(Z-1)$, not $Z$, that appears: the electron dropping into the K shell “sees” the nucleus screened by the electron that remains there.

$\lambda=\dfrac{1240}{8000}=0.155$ nm, the line used in laboratory diffractometers.

$E=\dfrac{1240}{0.194}=6.39\times 10^{3}$ eV, so $(Z-1)^{2}=\dfrac{6392}{10.2}=627$, $Z-1=25.0$ and $Z=26$: IRON. This is how Moseley ordered the elements by atomic number and predicted the missing elements of the periodic table.

As long as its energy stays below $4.9$ eV, the electron bounces off the mercury atoms without losing energy (elastic collisions). As soon as it reaches $4.9$ eV, it can give that energy to an atom by raising it to its first excited state: it loses almost all of its speed and no longer gets past the retarding grid, hence the drop in current.

The excited atom returns to the ground state by emitting a photon of $4.9$ eV: $\lambda=\dfrac{1240}{4.9}=253$ nm, in the ultraviolet. This is indeed the observed line, the same one as in germicidal mercury lamps.

An electron can excite an atom, be accelerated again, excite a second one, and so on: drops at $4.9$ V, $9.8$ V and $14.7$ V. It is the constant spacing between the peaks, not the position of the first one (shifted in practice by the difference in the work functions of the electrodes), that measures the excitation energy.

$v=\sqrt{\dfrac{2E}{m_{e}}}=\sqrt{\dfrac{2\times 4.9\times 1.60\times 10^{-19}}{9.11\times 10^{-31}}}=1.31\times 10^{6}$ m/s.

It shows that the atom accepts energy only in definite packets, even when that energy comes from a collision with an electron: quantization belongs to the ATOM, not only to light. And the emitted photon has exactly the energy measured electrically, which links the two methods.

Intensity sets the NUMBER of photons per second; the energy of each photon depends only on its frequency, $E=hf$. A powerful red laser sends many photons of $1.9$ eV, a weak ultraviolet lamp few photons of $4$ eV.

This is the Bohr model, useful for energies but superseded. Quantum mechanics gives no trajectory: it describes the electron by a wave whose square gives a PROBABILITY of presence, spread over “orbitals” with no sharp outline. The Bohr radius remains the most probable distance in the ground state.

Absorption occurs only for energies equal to a gap between levels: $10.2$ eV ($1\rightarrow 2$), $12.09$ eV ($1\rightarrow 3$), $12.75$ eV ($1\rightarrow 4$)... Above $13.6$ eV, ionization accepts any energy. For $12.75$ eV: $E_{n}=-13.6+12.75=-0.85$ eV, so $n^{2}=\dfrac{13.6}{0.85}=16$ and $n=4$.

Light does not change nature: it is a single quantum object, whose propagation is calculated like a wave and whose exchanges of energy take place in quanta. Sent photon by photon through Young's slits, it strikes the screen in point-like impacts that, in the long run, draw the interference pattern: the two aspects appear in the SAME experiment.

They fall at the SAME wavelengths, because they correspond to the same gaps between energy levels: the photon absorbed to go up from $E_{1}$ to $E_{2}$ has the energy of the one emitted on the way back down. This is what makes it possible to identify the elements of a star from the dark lines of its spectrum.

The observed wavelength is LONGER: a redshift, so the galaxy is receding.

$v=c\dfrac{\Delta\lambda}{\lambda}=3.00\times 10^{5}\times\dfrac{2.1}{656.3}=9.6\times 10^{2}$ km/s. We check that $v\ll c$ ($0.3$ %): the non-relativistic formula is valid.

$d=\dfrac{v}{H_{0}}=\dfrac{960}{70}=13.7$ Mpc, about $45$ million light-years.

The RELATIVE shift is the same for all lines: $\lambda=486.1\times\left(1+\dfrac{2.1}{656.3}\right)=487.7$ nm. The absolute shift, $1.6$ nm, is smaller than that of H$\alpha$: this is the signature of a Doppler effect, and not of a calibration error, which would shift everything by the same amount.

Because ALL the lines of hydrogen (H$\alpha$, H$\beta$, H$\gamma$...) are shifted by the same factor $1+\dfrac{v}{c}$: the pattern of wavelength ratios set by the Bohr levels is preserved. We recognize the fingerprint of the element, then measure its shift.

Area of the pupil: $\pi\times(3.5\times 10^{-3})^{2}=3.85\times 10^{-5}$ m². Power: $P=1.0\times 10^{-11}\times 3.85\times 10^{-5}=3.85\times 10^{-16}$ W. The RADIUS is half the diameter.

$E=\dfrac{hc}{\lambda}=\dfrac{6.63\times 10^{-34}\times 3.00\times 10^{8}}{550\times 10^{-9}}=3.62\times 10^{-19}$ J.

$N=\dfrac{P}{E}=\dfrac{3.85\times 10^{-16}}{3.62\times 10^{-19}}=1.06\times 10^{3}$ photons/s.

$1.06\times 10^{3}\times 0.10\approx 106$ photons, of which only a small fraction is absorbed by the rod cells.

A hundred times less light means a hundred times fewer photons: $\dfrac{1.06\times 10^{3}}{100}\approx 11$ photons per second, one about every $94$ ms, so $11\times 0.10\approx 1$ photon per integration time. The brain reports a light only if several neighbouring rod cells are triggered within the same time interval: this is a filter against the “noise” of spontaneous triggering. With a single photon per integration time, most often lost before reaching a rod cell, the threshold is no longer crossed. Light remains granular at this level: a more sensitive detector, such as a photon-counting camera, would see it arrive grain by grain.

$\lambda_{n}=\dfrac{2L}{n}$, so $p_{n}=\dfrac{h}{\lambda_{n}}=\dfrac{nh}{2L}$ and $E_{n}=\dfrac{p_{n}^{2}}{2m_{e}}=\dfrac{n^{2}h^{2}}{8m_{e}L^{2}}$. For $L=1.00\times 10^{-9}$ m: $E_{1}=\dfrac{(6.63\times 10^{-34})^{2}}{8\times 9.11\times 10^{-31}\times 10^{-18}}=6.03\times 10^{-20}$ J $=0.377$ eV. Quantization comes here from the standing-wave condition alone, with no additional postulate.

$E_{2}-E_{1}=(4-1)E_{1}=3\times 0.377=1.13$ eV.

$\lambda=\dfrac{1240}{1.131}=1.10\times 10^{3}$ nm, in the near infrared.

For $L=0.73$ nm, $E_{1}=\dfrac{0.377}{0.73^{2}}=0.707$ eV. The $6$ electrons fill levels $1$, $2$ and $3$; the transition $3\rightarrow 4$ costs $(16-9)E_{1}=7\times 0.707=4.95$ eV, that is $\lambda=\dfrac{1240}{4.95}=250$ nm. The measured value is $258$ nm: a one-line model recovers the absorption of a real molecule to within $3$ %. The longer the chain, the larger $L$ becomes and the more the absorbed colour slides toward the visible, hence the orange colour of carotene.

The level $n=0$ would give an infinite $\lambda$, a wave with no interior node that does not vanish at the walls: this is forbidden. The minimum energy is $E_{1}>0$, the “zero-point energy”, which grows as $L$ decreases. This is the uncertainty principle at work: to confine is to give momentum.

$\dfrac{g_{2}}{g_{1}}=\dfrac{2\times 4}{2\times 1}=4$.

$kT=8.62\times 10^{-5}\times 5800=0.500$ eV, twenty times less than the gap of $10.2$ eV.

$\dfrac{N_{2}}{N_{1}}=4\,e^{-10.2/0.500}=4\,e^{-20.4}=5.5\times 10^{-9}$. Not even one atom in a hundred million: the Sun nevertheless has visible Balmer lines, because the column of gas it passes through contains an enormous number of atoms.

$kT=0.862$ eV and $\dfrac{N_{2}}{N_{1}}=4\,e^{-11.83}=2.9\times 10^{-5}$: five thousand times more than at $5800$ K, for a temperature barely doubled. The exponential makes the population of level $2$ extraordinarily sensitive to temperature.

Beyond roughly $10\,000$ K, hydrogen becomes massively IONIZED: the ionization energy is only $13.6$ eV, and atoms that have lost their electron can no longer absorb a line. The population in level $2$ therefore goes through a maximum. This is how Cecilia Payne showed, in 1925, that the differences between stellar spectra come from temperature and not from composition.