The box of an LED bulb in any Indian shop says "9 W, as bright as 60 W". A good part of that power leaves the bulb as light, as photons. So the bulb sends out some definite number of photons every second. Written out, that number has nineteen digits. Yet the room looks evenly lit, with no flicker of separate packets at all.
A particle with no mass at rest
The photon behaves like a particle in one sense: it is absorbed whole, at one place, in one event. So it is fair to ask what else a particle has. A photon always moves at the speed of light, \(c = 3 \times 10^{8}\,\text{m/s}\) in vacuum. It can never be slowed down or brought to rest. Its mass when at rest, its rest mass, is zero.
A body with zero rest mass still carries energy and momentum. The theory of relativity, in the next chapter, shows that for such a body \(E = pc\). With \(E = h\nu\) and \(c = \nu\lambda\), this gives \(p = h\nu/c = h/\lambda\). The shorter the wavelength, the larger the momentum.
Notice what the equation \(p = h/\lambda\) joins together. Momentum is a property of particles. Wavelength is a property of waves. One constant, \(h\), links the two. The next chapter turns this equation round and gives a wavelength to the electron.
Find the energy and the momentum of a photon of green light of wavelength \(550\,\text{nm}\).
(a) \(E = hc/\lambda = 6.63 \times 10^{-34} \times 3 \times 10^{8} / 550 \times 10^{-9} = 3.62 \times 10^{-19}\,\text{J}\). In eV this is \(3.62 \times 10^{-19} / 1.6 \times 10^{-19} = 2.26\,\text{eV}\).
(b) \(p = h/\lambda = 6.63 \times 10^{-34} / 550 \times 10^{-9} = 1.21 \times 10^{-27}\,\text{kg m/s}\).
(c) Check: \(p = E/c = 3.62 \times 10^{-19} / 3 \times 10^{8} = 1.21 \times 10^{-27}\,\text{kg m/s}\).
Values used: \(h = 6.63 \times 10^{-34}\,\text{J s}\), \(c = 3 \times 10^{8}\,\text{m/s}\), \(e = 1.6 \times 10^{-19}\,\text{C}\).
The shortcut \(1240/550\) gives \(2.25\,\text{eV}\), while \(6.63 \times 10^{-34} \times 3 \times 10^{8}/(1.6 \times 10^{-19})\) is really \(1243\,\text{eV nm}\) and gives \(2.26\,\text{eV}\). The difference comes only from rounding the constants. Use one route in one problem, and both answers are accepted.
A gamma-ray photon and a radio photon travel through empty space. Which statement is correct?
Find the momentum of a photon of red light of wavelength \(663\,\text{nm}\). Give the answer in units of \(10^{-27}\,\text{kg m/s}\).
In 1923 Arthur Compton scattered X-ray photons off electrons and found them coming out with longer wavelengths. The shift fitted a collision in which the photon carries momentum \(h/\lambda\) and gives some of it to the electron, like one ball striking another.
Counting photons
A source of light power \(P\) sends out \(P\) joules of light every second. Each photon carries \(E = hc/\lambda\). So the number of photons it sends each second, its photon rate \(n\), is the power divided by the energy of one photon.
Now the LED box has its answer. Suppose the bulb gives out \(3\,\text{W}\) of light, taken as green light of \(550\,\text{nm}\). Each photon carries \(3.62 \times 10^{-19}\,\text{J}\). So the bulb sends out \(3/3.62 \times 10^{-19} = 8.3 \times 10^{18}\) photons every second. That is 8 300 000 000 000 000 000, nineteen digits.
A small source sends its photons equally in all directions. At a distance \(r\) they cross a sphere of area \(4\pi r^2\), just as the intensity of sound did in the chapter on loudness. So the photons crossing each square metre per second are \(n/4\pi r^2\). They thin out as \(1/r^2\).
A small lamp gives out \(12\,\text{W}\) of light of wavelength \(600\,\text{nm}\), equally in all directions. How many photons does it emit each second? How many cross each square metre each second at \(5\,\text{m}\)?
One photon carries \(E = hc/\lambda\).
\(n = P/E\), and at distance \(r\) the rate per square metre is \(n/4\pi r^2\).
\(E = 6.63 \times 10^{-34} \times 3 \times 10^{8} / 600 \times 10^{-9} = 3.32 \times 10^{-19}\,\text{J}\), so \(n = 12/3.32 \times 10^{-19} = 3.6 \times 10^{19}\) per second. At \(5\,\text{m}\), \(3.6 \times 10^{19}/(4\pi \times 25) = 1.15 \times 10^{17}\) per square metre per second.
Even five metres away, about \(10^{13}\) photons cross each square centimetre of a page every second.
A laser pointer gives out \(1.0\,\text{mW}\) of red light of wavelength \(660\,\text{nm}\). How many photons does it emit each second? Answer in units of \(10^{15}\) per second.
Source A gives out \(1\,\text{W}\) of light of \(620\,\text{nm}\), and source B gives out \(1\,\text{W}\) of \(310\,\text{nm}\). Find the ratio of the photon rates \(n_A : n_B\).
Brighter means more, not bigger
In the photon picture, the intensity of a beam is its photon rate per square metre times the energy of one photon. Doubling the power of a green lamp doubles the number of green photons. Each photon stays exactly the same size, \(2.26\,\text{eV}\). To make bigger photons, you must change the colour.
The numbers explain why light looks smooth. Even a faint lamp sends out billions of billions of photons each second. The eye and the photocell see only their average flow, just as a bucket filling under a tap never shows the separate water molecules. In very dim light the lumps begin to show.
Photons in, electrons out
A photocell never turns every photon into an electron. Many photons are reflected, and others give their energy to electrons deep in the metal that never escape. The fraction of photons that free an electron is the quantum efficiency of the surface. It is often a few per cent. The saturation current counts the electrons, so \(I = (\text{efficiency}) \times n \times e\).
A beam sends \(10^{16}\) photons each second to a metal plate. One photon in every hundred frees a photoelectron, and all of them reach the collector. What is the saturation current, in microamperes?
The power of a lamp of one fixed colour is doubled. Which statements are true?
A violet photon has a wavelength of \(400\,\text{nm}\), and an FM radio photon a wavelength of \(3\,\text{m}\). How many radio photons carry the same energy as one violet photon? Answer in millions.
Each pixel of a phone camera counts the photons that land on it during the exposure. In daylight a pixel collects many thousands, and the random scatter in the count is tiny beside the count. At night a pixel may catch only a few dozen photons. The count then varies noticeably from pixel to pixel, even where the scene is evenly dark. That random speckle is the grain of a night photograph. It is the lumpiness of light made visible.
- Thinking that more energetic photons travel faster.Every photon in vacuum moves at \(c\). Energy changes the frequency and the momentum, never the speed.
- Using the electrical power drawn by a bulb as its light power.\(n = P/E\) needs the power that leaves as light, which is only part of the power drawn.
- Assuming equal powers mean equal numbers of photons.\(n = P\lambda/hc\). At equal power, the longer wavelength sends more photons.
- Forgetting the quantum efficiency when finding a current from photons.Only a fraction of the photons free electrons. Multiply the photon rate by that fraction, then by \(e\).