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Physics · Ch 7 — Dual Nature of Radiation and Matter

Particle nature of light: Einstein's explanation

7.2.7

Particle nature of light: Einstein's explanation

In 1905, Einstein extended Planck's quantum idea from the atomic oscillators that emit or absorb light to light itself. Rather than picturing the energy in a light wave as spread out continuously over its wavefront, Einstein proposed that light of frequency ν\nu from any source should be thought of as a stream of discrete energy packets, or quanta, each carrying energy E=hνE=h\nu. Each such quantum of light behaves, in every relevant respect, like a genuine particle, and is called a photon. Einstein further proposed that a photon carries a definite linear momentum of magnitude p=hν/cp=h\nu/c.

From this single reconception of light, the following characteristic properties of photons follow. (i) A photon of frequency ν\nu and wavelength λ\lambda carries energy E=hν=hc/λE=h\nu=hc/\lambda. (ii) A photon's energy is set entirely by the frequency of the radiation, never by its intensity -- intensity instead reflects only how many photons per second are arriving, with each individual photon's energy unaffected. (iii) Photons travel at the speed of light and carry momentum p=hν/c=h/λp=h\nu/c=h/\lambda. (iv) Being electrically neutral, photons are completely unaffected by electric and magnetic fields. (v) In any photon-electron interaction, total energy, total linear momentum and total angular momentum are all conserved, though the number of photons itself need not be conserved, since a photon can be absorbed entirely or a new one produced in the interaction.

Applying the photon picture to the photoelectric effect: when a single photon of energy hνh\nu strikes a metal surface, it is completely absorbed by one single electron. Part of this absorbed energy, equal to the photoelectric work function ϕ0\phi_0, is used up ejecting the electron from the metal, and the remainder becomes the ejected electron's kinetic energy. Conservation of energy for this one photon-one electron interaction gives Einstein's photoelectric equation,

hν=ϕ0+12mv2(7.6)h\nu=\phi_0+\tfrac12 mv^2 \qquad (7.6)

If an electron loses no energy to internal collisions on its way out, it emerges with the maximum possible kinetic energy Kmax=12mvmax2K_{max}=\tfrac12 m v_{max}^2, and equation (7.6) rearranges to the most commonly used form,

Kmax=hν−ϕ0(7.8)K_{max}=h\nu-\phi_0 \qquad (7.8)

At the threshold frequency ν0\nu_0, the photoelectron is emitted with essentially zero kinetic energy, so equation (7.6) reduces to hν0=ϕ0h\nu_0=\phi_0 (7.7 combines these as hν=hν0+12mv2h\nu=h\nu_0+\tfrac12mv^2). Plotting KmaxK_{max} against ν\nu from equation (7.8) gives a straight line whose slope is hh and whose yy-intercept, extended back to zero frequency, is −ϕ0-\phi_0; this is exactly the graph R. A. Millikan produced experimentally for several metals (caesium, potassium, sodium, calcium), finding that although each metal's line sits at its own characteristic height (fixed by that metal's own work function), every line has the identical slope hh -- Millikan's precise experimental route to Planck's constant, h=6.626×10−34h=6.626\times10^{-34} Js.

Einstein's equation explains every one of the five experimental laws at once. (i) Since each absorbed photon liberates exactly one electron, more photons per second (higher intensity) simply means more electrons emitted per second, i.e. more photocurrent -- exactly as observed. (ii) From Kmax=hν−ϕ0K_{max}=h\nu-\phi_0, the maximum kinetic energy depends only on frequency, never on intensity -- resolving the first puzzle from section 7.2.6. (iii) Since a photon needs at least energy ϕ0\phi_0 to free an electron at all, there must exist a minimum (threshold) frequency ν0=ϕ0/h\nu_0=\phi_0/h below which no photoelectron can ever be emitted, however many photons arrive -- resolving the second puzzle. (iv) Because each photon's energy is transferred to a single electron in one complete, instantaneous absorption event rather than being slowly accumulated, there is no time lag between the light striking the surface and the electron being ejected -- resolving the third puzzle. …

Figure 7.13Emission of photoelectrons

What this figure shows. Two small panels each show a metal surface being struck by an incoming photon of energy E=hνE=h\nu, with an electron shown leaving the surface in panel (a) carrying maximum kinetic energy Kmax=hν−hν0K_{max}=h\nu-h\nu_0, corresponding to light above the threshold frequency, while panel (b) shows the special case at exactly the threshold frequency ν0\nu_0, where the photon's energy E=hν0E=h\nu_0 is just enough to free the electron and it leaves the surface with essentially zero kinetic energy, K=0K=0. The two panels together are the pictorial version of Einstein's photoelectric equation: as the incident frequency is reduced toward ν0\nu_0, the emitted electron's kin …

Figure 7.14Kmax vs ν graph

What this figure shows. A graph with frequency ν\nu of the incident light along the horizontal axis and maximum kinetic energy KmaxK_{max} of the photoelectrons along the vertical axis, drawn as a single straight line that crosses the frequency axis at the threshold frequency ν0\nu_0 and has a y-intercept of −ϕ0-\phi_0 when extended back to zero frequency. The line's slope is explicitly labelled hh -- Planck's constant -- making this graph the direct experimental route to measuring hh: it is exactly the straight-line form Kmax=hν−ϕ0K_{max}=h\nu-\phi_0 that Einstei …

Figure 7.15Kmax vs ν graph for different metals

What this figure shows. The same kind of KmaxK_{max}-versus-ν\nu straight-line graph as Figure 7.14, but now drawn simultaneously for four different metals -- caesium, potassium, sodium and calcium -- as four parallel lines, each starting from its own y-intercept marked with that metal's own negative work function value (-2.14 eV, -2.30 eV, -2.75 eV and -3.20 eV respectively) but all sharing the identical slope, explicitly labelled 'Slope = h'. This is exactly the graph Robert Millikan produced to experimentally verify Einstein's equation: the four lines' common slope, independent of which metal is used, gave Millikan his precise measurement of Planck's constant, h=6.626×10−34h=6.626\times10^{-34} Js, wh …

Misc Example 7.2Will 300 nm radiation eject photoelectrons from silver?

Worked out. A radiation of wavelength 300 nm is incident on a silver surface, and the task is to determine whether photoelectrons will be observed at all. Computing the incident photon's energy using E=hc/λE=hc/\lambda (expressed in eV) with the standard constants gives E≈4.14E\approx4.14 eV. Comparing this against silver's work function, listed in Table 7.1 as 4.7 eV, shows the photon energy (4.14 eV) is less than the work function (4.7 eV) required to free even the most weakly bound surface electron -- so no photoelectrons are observed for this radiation on silver, however long or brightly it is shone, illustrating the strict quantum threshold-energy condition that a …

Misc Example 7.3Threshold wavelength and stopping potential for copper at 2200 Å

Worked out. When light of wavelength 2200 Å falls on copper (work function ϕ0=4.65\phi_0=4.65 eV), photoelectrons are emitted, and the task is to find both the threshold wavelength and the stopping potential. The threshold wavelength follows directly from λ0=hc/ϕ0\lambda_0=hc/\phi_0, giving λ0≈2672\lambda_0\approx2672 Å -- the longest wavelength (lowest frequency) copper can still respond to. The incident photon's own energy at 2200 Å works out to E=hc/λ≈9.035×10−19E=hc/\lambda\approx9.035\times10^{-19} J ≈5.65\approx5.65 eV, so applying Einstein's equation gives the photoelectrons' maximum kinetic energy as Kmax=hν−ϕ0=5.65−4.65=1K_{max}=h\nu-\phi_0=5.65-4.65=1 eV. Since Kmax=eV0K_{max}=eV_0, the corresponding stopping potential is simply V0=1V_0=1 V -- a clean numerical illustration of how a single incident wavelength, together with a metal's tabulated work function, pins down both the thresho …

Misc Example 7.4Photoelectrons from potassium under 3000 Å UV light

Worked out. UV light of wavelength 3000 Å and intensity 2 W m−22\ \text{W m}^{-2} is incident on a potassium surface (work function 2.30 eV) of area 2 cm22\ \text{cm}^2, and the task has two parts. First, the incident photon's energy works out to E=hc/λ≈6.626×10−19E=hc/\lambda\approx6.626\times10^{-19} J ≈4.14\approx4.14 eV, so by Einstein's equation the maximum kinetic energy of the ejected photoelectrons is Kmax=hν−ϕ0=4.14−2.30=1.84K_{max}=h\nu-\phi_0=4.14-2.30=1.84 eV. Second, assuming 40% of the incident photons actually succeed in producing a photoelectron, the number of photons striking the surface per second is found from the incident power divided by the single-photon energy, np=IA/E≈6.04×1014n_p=IA/E\approx6.04\times10^{14} photons per second, so the rate of photoelectron emission is 0.40×np≈2.415×10140.40\times n_p\approx2.415\times10^{14} photoelectrons per second -- showing how a photon-counting argument …

Misc Example 7.5Work function and threshold wavelength of a metal from a stopping-potential measurement

Worked out. Light of wavelength 390 nm directed at a metal electrode is found to be completely stopped by an opposing potential difference of 1.10 V, and the task is to find both the metal's work function and its threshold wavelength for ejecting electrons. Since Kmax=eV0K_{max}=eV_0 at the stopping potential, Einstein's equation ϕ0=hν−Kmax\phi_0=h\nu-K_{max} can be rewritten as ϕ0=hcλ−eV0\phi_0=\dfrac{hc}{\lambda}-eV_0; substituting the given wavelength and stopping potential gives ϕ0≈3.34×10−19\phi_0\approx3.34\times10^{-19} J ≈2.09\approx2.09 eV. The threshold wavelength then follows from λ0=hc/ϕ0\lambda_0=hc/\phi_0, giving λ0≈5969\lambda_0\approx5969 Å (about 5963-5969 Å depending on rounding) -- demonstrating that a single stopping-potential measurement at one known wavelength is, by itself, enough to fully characterise …