Light and electrons as waves and particles
Why does very weak light or an electron beam draw first random single points on the screen, from which a fringe pattern arises only after thousands of hits?
Prior knowledge
- Two waves strengthen or weaken each other depending on the path difference between themchapter Interference of waves
- The energy of a photon is $E = hf = \tfrac{hc}{\lambda}$, and one photon can release at most one electron from a metalchapter Photoelectric effect and the photon
- When an atom moves to a lower energy level, it emits a photon whose energy is the difference between the levelschapter Energy levels of the hydrogen atom and the Bohr model
Core path: follow the steps in order. Tasks marked Optional give extra practice. Optional tasks: 2.
Think about it
In 1989 a Japanese research group sent electrons one at a time towards a device equivalent to a double slit and filmed the screen. The first electrons appeared as single points of light that flashed up at seemingly random places. When tens of thousands of points had accumulated, regular fringes could be seen on the screen. The same experiment can be done with light so weak that there is usually only one photon in the apparatus at a time.
Does each electron or photon pass through only one of the slits, and if it does, where do the fringes come from?
Review: the photoelectric effect
Predict
Make your predictions before investigating 2 questions
Prediction 1: the second slit open
Prediction 2: measuring the path
Investigate
Send photons or electrons one at a time through two slits and see how a pattern arises from single hits.
Simulation
How to use
Start sends quanta one at a time: a faint wave spreads from the source through the slits, and the quantum is detected on the screen as a single point. +1000 hits adds a thousand hits at once. Clear empties the screen. Choose the Particles, the Slits (only left, both or only right) and the Path measurement. When the measurement is on, the lamp next to the slits lights up for the slit the quantum passed through. At the top is the apparatus seen from above, in the middle the screen seen from the front, and at the bottom a bar chart of the hits in $0.5\,\text{mm}$ intervals. The slider moves the green counting interval $x_0 \pm 0.2\,\text{mm}$. When you change the particles, the slits or the measurement, the screen is cleared. Speed changes only the pace of the animation.
Assumptions of the model
The hit positions are drawn from the probability distribution of the model, so the numbers are values of the model, not measurements. The drawing starts from the same point after every clearing, so the same experiment always gives the same numbers. The patterns of photons and electrons are drawn to the same scale to make comparison easier; in real experiments the fringe spacing is quite different. The width and the distance of the slits are drawn enlarged. The path measurement is ideal: it always tells the right slit. There is only one quantum in the apparatus at a time, and no quanta are lost on the way. The faint wave illustrates the probability wave and is not a measurable quantity.
How the calculation works
When both slits are open and the path is not measured, the probability of hits at the point $x$ is proportional to the intensity of the wave model
$$I(x) \propto \cos^2\!\left(\frac{\pi x}{s}\right) \cdot \left(\frac{\sin(\pi x / a)}{\pi x / a}\right)^2$$
where the fringe spacing is $s = 2.0\,\text{mm}$ and the first zero of the one-slit pattern is at $a = 10\,\text{mm}$. With one slit or a measured path the first factor is missing, so there are no fringes. A dark fringe is where the $\cos^2$ factor is zero: $x = \pm 1.0\,\text{mm}, \pm 3.0\,\text{mm}, \ldots$. The share is $n : N$, where $n$ is the number of hits in the counting interval and $N$ the total number of hits. The dashed line is the expected number of hits: $N$ times the probability of the interval.
Experiment 1: one slit or two
Test prediction 1. The photons, no path measurement and the counting interval $x_0 = 1.0\,\text{mm}$ stay constant; change only the slits. First follow a few single photons with the Start button. Then press +1000 hits and record the measurement. Open both slits, press +1000 hits again and record.
Initial state: particles: photons, slits: only left, path measurement: not measured, counting interval = $1\,\text{mm}$. Kept constant: particles, path measurement, counting interval. Varied: slits.
| slits | $N$ | $n$ | $n$ : $N$ $(\%)$ |
|---|
Experiment 2: through which slit?
Test prediction 2. Now electrons are sent, and both slits are open. The counting interval is in the dark fringe $x_0 = 1.0\,\text{mm}$. Press +1000 hits without the path measurement and record. Then switch the measurement on, press +1000 hits and record. Compare the patterns on the screen.
Initial state: particles: electrons, slits: both, path measurement: not measured, counting interval = $1\,\text{mm}$. Kept constant: particles, slits, counting interval. Varied: path measurement.
| path measurement | $N$ | $n$ | $n$ : $N$ $(\%)$ |
|---|
Conclusion: bullet or wave?
Write, based on your tables: how did the share in the counting interval change when the second slit was opened, and when the path was measured? Why can the result not be explained by each quantum passing through one slit like a bullet? What is the connection between the wave and a single hit?
Model answer
With one slit, $43$ photons out of $1000$ hit the counting interval $x_0 = 1.0\,\text{mm}$, but when both slits were open, only $3$. If the photons passed through one of the slits as bullets, the distribution would be the sum of two one-slit patterns, and the share would stay the same as with one slit. Both slits affect every quantum: its wave passes through both slits, and at this point the waves weaken each other. The electrons behaved in the same way. When the path was measured, the dark fringe disappeared ($43$ hits), because information about the path removes the interference. A single hit is a random point, but the intensity of the wave tells the probability of a hit at each place. The numbers are values of the model.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Compton's scattering experiment
Values calculated by a model
In 1923 Arthur Compton directed X-rays at graphite and measured the wavelength $\lambda'$ of the radiation scattered in different directions. The wavelength of the incident radiation was $\lambda = 0.0711\,\text{nm}$. The values in the table have been calculated from Compton's model, and Compton's measurements followed them.
| Scattering angle | $\lambda'$ ($\text{nm}$) |
|---|---|
| $0^\circ$ | $0.0711$ |
| $45^\circ$ | $0.0718$ |
| $90^\circ$ | $0.0735$ |
| $135^\circ$ | $0.0752$ |
| $180^\circ$ | $0.0760$ |
Theory summary
Key idea
Wave or particle?
In the double-slit experiment, light passes through two narrow, closely spaced slits to a screen. A pattern of bright and dark fringes arises on the screen, which the wave model explains: the waves from the two slits strengthen each other where the path difference is a multiple of the wavelength, and weaken each other where the path difference is half a wavelength, one and a half wavelengths or another odd multiple of half a wavelength (chapter Interference of waves). In the photoelectric effect, however, light gives up its energy in quanta: one photon, whose energy is $hf$, can release at most one electron (chapter Photoelectric effect and the photon). Both observations are true. How can the same light be both?
One quantum at a time
When the double-slit experiment is done with light so weak that there is on average less than one photon in the apparatus at a time, a sensitive detector registers each photon as a single point. The position of a single point cannot be predicted. When thousands of points have accumulated, they form the same fringe pattern as strong light. The same happens with electrons: in a double-slit experiment with an electron beam, each electron hits the screen at one place, but the hits pile up into fringes. So the electron, which we are used to thinking of as a small particle, behaves like a wave as it travels, and the photon, which the wave model describes well, is detected like a particle.
The wave gives the probability
The wave model does not tell where a single photon or electron will hit. It tells the probability with which the quantum is detected at each place: where the intensity of the wave is large, many hits arrive, and in the dark fringes hardly any. When only one slit is open, the fringes of the two slits disappear. When both slits are open, fewer hits arrive in the dark fringes than with one slit, so opening the second slit reduces the hits at these places. This cannot be explained by each quantum passing through only one slit like a small bullet. If the experiment measures through which slit each quantum passes, the fringe pattern disappears and the distribution is the sum of two one-slit patterns.
The phenomenon in which the same object has both wave and particle properties is called wave-particle duality. It does not mean that a photon or an electron changes its form. Neither classical model, wave or bullet, describes a quantum object alone, and the experimental arrangement decides which property shows. While travelling, a quantum object behaves like a wave, and when it is detected, it is registered at one place like a particle.
Four models of the photon
The photon is described in physics with several models. Each works in its own range of phenomena, and none of them alone is the whole picture.
- State of the field: light is an oscillation of the electric and magnetic field that propagates as a wave. The model explains interference and diffraction.
- Quantum of the field: the energy of the radiation field changes only in whole quanta $hf$. The model explains the photoelectric effect and spectral lines.
- Quantum particle: the photon is detected at one place, and it carries energy and momentum. The model explains single hits and the Compton effect.
- Mediator of an interaction: the electromagnetic interaction between electric charges is mediated by photons. The model belongs to the Standard Model of particle physics.
A photon has no rest mass, and in a vacuum it always travels at the speed of light $c$. So it is not a small bullet that could be at rest.
Photon momentum
Although a photon has no mass, it has a momentum that depends on its energy:
$$p = \frac{E}{c} = \frac{h}{\lambda}$$
where $p$ is the momentum of the photon ($\text{kg}\cdot\text{m/s}$), $E = hf = \tfrac{hc}{\lambda}$ the energy of the photon ($\text{J}$), $c = 2.998 \cdot 10^8\,\text{m/s}$ the speed of light, $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$ the Planck constant and $\lambda$ the wavelength ($\text{m}$). The unit works out: $\tfrac{\text{J}\,\text{s}}{\text{m}} = \tfrac{\text{kg}\cdot\text{m}^2/\text{s}^2 \cdot \text{s}}{\text{m}} = \text{kg}\cdot\text{m/s}$. The momentum of a photon of visible light is small: at the wavelength $500\,\text{nm}$ it is ${1.3 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}}$. The momentum of an X-ray photon of wavelength $0.0711\,\text{nm}$ is about $7000$ times as large, ${9.3 \cdot 10^{-24}\,\text{kg}\cdot\text{m/s}}$. A short wavelength thus means a large energy and a large momentum.
The Compton effect
In 1923 Arthur Compton directed X-rays at graphite and measured the wavelength of the scattered radiation. The wavelength of the scattered radiation was longer than that of the incident radiation, and the difference grew as the scattering angle grew. This is called the Compton effect. The wave model does not explain it: electrons oscillating in time with the wave would radiate at the same frequency as the incident radiation. In the photon model the phenomenon is a collision. The photon gives part of its energy and momentum to an almost free electron, which starts to move. Energy is conserved:
$$hf = hf' + E_\text{k}$$
where $hf$ is the energy of the incident photon, $hf'$ the energy of the scattered photon and $E_\text{k}$ the kinetic energy received by the electron. Because $hf' < hf$, the frequency of the scattered photon is lower and its wavelength $\lambda'$ longer than those of the incident one. Momentum is also conserved as a vector quantity: $\overline{p} = \overline{p}' + \overline{p}_e$, where $\overline{p}_e$ is the momentum of the electron.
Photon flow
The power $P$ of light tells how much energy the light carries per second. Because each photon brings the energy $hf$, the number of photons passing per second is
$$\frac{N}{t} = \frac{P}{hf}$$
where $N$ is the number of photons, $t$ the time ($\text{s}$), $P$ the power ($\text{W}$) and $hf$ the energy of one photon ($\text{J}$). In ordinary light there are so many photons that the light looks like a continuous stream. Only in very weak light do single photons stand out.
Common misconception
Many people think that the fringe pattern of the double-slit experiment arises when photons or electrons collide with each other, or that each quantum passes through just one of the slits like a small bullet. In the weak light of the example the photons arrive one at a time, and the pattern still arises. When the same total number of hits is counted, in the bullet model the share $n : N$ of hits at a given place would stay the same as with one slit, because the second slit would only add hits (if the source emits the same number of quanta per second for the same time). In reality the share of hits in the dark fringes is clearly smaller than with one slit.
Another mistake is to think that a photon has no momentum because it has no mass. In the Compton effect the photon gives the electron both energy and momentum, and light that is reflected from a mirror or absorbed by a surface pushes on it. The momentum of a photon is $p = \tfrac{h}{\lambda}$.
A third mistake is to think that wave-particle duality means that the object is sometimes a wave and sometimes a particle. A quantum object is the same all the time: the wave model gives the probability of detection, and when it is detected, the quantum is registered at one place.
Summary
- In the double-slit experiment, single photons and electrons hit the screen one at a time at random places, but thousands of hits form the fringe pattern of the wave model.
- The wave gives the probability: where the intensity of the wave is large, many hits arrive. If it is measured through which slit the quantum passes, the fringes disappear.
- Wave-particle duality: photons and electrons have both wave and particle properties, and the experimental arrangement decides which shows.
- The photon is described by four models: state of the field, quantum of the field, quantum particle and mediator of an interaction.
- The momentum of a photon is $p = \tfrac{E}{c} = \tfrac{h}{\lambda}$. In the Compton effect the photon gives the electron energy and momentum, so the wavelength of the scattered photon is longer: $hf = hf' + E_\text{k}$.
- The number of photons per second is $\tfrac{N}{t} = \tfrac{P}{hf}$.
Optional Return to your predictions
Compare your predictions with the values of the model in the experiments. Then return to the question at the beginning: does each electron pass through only one slit, and where do the fringes come from?
Would you change your prediction?
Apply
Worked example
Problem
In a double-slit experiment, red laser light of wavelength $633\,\text{nm}$ is used. The light is dimmed with filters so that the power of the light arriving at the slits is $1.0 \cdot 10^{-13}\,\text{W}$. How many photons arrive at the slits per second, and how long is the distance between successive photons on average? Is there usually more than one photon at a time in an apparatus of length $1.2\,\text{m}$?
Step 1/7
Given and asked. ${\lambda = 633\,\text{nm}}$, ${P = 1.0 \cdot 10^{-13}\,\text{W}}$, ${h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}$ and ${c = 2.998 \cdot 10^8\,\text{m/s}}$. We are asked for the number of photons per second $\tfrac{N}{t}$ and the average distance $d$ between successive photons.
Step 2/7
Principle. The energy of light travels as photons, so the power is the total energy of the photons arriving per second: $P = \tfrac{N}{t} \cdot hf$. The photons travel at the speed $c$. In one second the light advances the distance $c$, and $\tfrac{N}{t}$ photons fit in this distance, so the average gap is the distance divided by the number of photons.
Step 3/7
Formula solved.
$$\begin{aligned} \frac{N}{t} &= \frac{P}{hf}, \quad hf = \frac{hc}{\lambda} \\ d &= \frac{c}{N/t} \end{aligned}$$
Step 4/7
Unit conversions. The wavelength in metres: $633\,\text{nm} = 633 \cdot 10^{-9}\,\text{m}$. The power is already in watts, that is, joules per second.
Step 5/7
Substitution. First we calculate the energy of one photon, then the number of photons per second and finally the gap between them:
$$\begin{aligned} hf &= \frac{hc}{\lambda} \\ &= \frac{6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}{633 \cdot 10^{-9}\,\text{m}} \\ &\quad \cdot 2.998 \cdot 10^{8}\,\tfrac{\text{m}}{\text{s}} \\ &= 3.138 \cdot 10^{-19}\,\text{J} \\[6pt] \frac{N}{t} &= \frac{P}{hf} \\ &= \frac{1.0 \cdot 10^{-13}\,\text{W}}{3.138 \cdot 10^{-19}\,\text{J}} \\ &= 3.187 \cdot 10^{5}\,\tfrac{1}{\text{s}} \\[6pt] d &= \frac{c}{N/t} \\ &= \frac{2.998 \cdot 10^{8}\,\text{m/s}}{3.187 \cdot 10^{5}\,\tfrac{1}{\text{s}}} \\ &= 940.8\,\text{m} \end{aligned}$$
Step 6/7
Answer. About $3.2 \cdot 10^5$ photons arrive at the slits per second, and the gap between successive photons is on average about $9.4 \cdot 10^2\,\text{m}$. Two significant figures, because the power $1.0 \cdot 10^{-13}\,\text{W}$ is known to two figures. The length of the apparatus, $1.2\,\text{m}$, is only about a thousandth ($0.13\,\%$) of the gap between the photons, so there is usually at most one photon in the apparatus at a time.
Step 7/7
Reasonableness. On average there are $1.2\,\text{m} : 940.8\,\text{m} \approx 0.0013$ photons in the apparatus, so most of the time there is no photon at all and two at once is rare. An ordinary $1\,\text{mW}$ laser pointer emits about $3 \cdot 10^{15}$ photons per second, so the light has been dimmed by a factor of about $10^{10}$, which dark filters can do. The fringe pattern therefore cannot arise from interaction between the photons.
Find the error
Apply to a new situation
A solar sail is a thin, mirror-like film that sunlight pushes in space. At the Earth's distance from the Sun, the intensity of sunlight, that is the power per area, is $1361\,\tfrac{\text{W}}{\text{m}^2}$. The area of the sail is $32\,\text{m}^2$, and it is perpendicular to the light. Assume that all the light is reflected straight back.
How large a force does the light exert on the sail? Give the answer in micronewtons ($\mu\text{N}$).
Hint 1: a guiding question
What is the connection between force and momentum? How much does the momentum of one photon change when it is reflected straight back?
Hint 2: what is the connection
Per second, the energy $P = IA$ arrives at the sail. The total momentum of all these photons is $\tfrac{P}{c}$, because for each photon $p = \tfrac{E}{c}$.
Hint 3: an intermediate step
In reflection the momentum changes twice as much:
Show the solution
Given
the intensity of sunlight $I = 1361\,\tfrac{\text{W}}{\text{m}^2}$, the area of the sail $A = 32\,\text{m}^2$, the light is reflected straight back; $c = 2.998 \cdot 10^8\,\text{m/s}$
Asked
the force $F$ exerted by the light on the sail in micronewtons
Principle
Force is the change of momentum per unit time. The momentum of each photon is $p = \tfrac{E}{c}$. When a photon is reflected straight back, its momentum changes from $p$ to $-p$, so the sail receives the momentum $2p$. Per second, the energy $P = IA$ arrives at the sail, so the momentum of the photons arriving per second is $\tfrac{P}{c}$.
Formula
where $P = IA$, so $F = \dfrac{2IA}{c}$
Conversions
Finally $1\,\text{N} = 10^6\,\mu\text{N}$.
Substitution
Answer
$2.9 \cdot 10^2\,\mu\text{N}$
Two significant figures as in the area $32\,\text{m}^2$.
Reasonableness
The force is about $0.3\,\text{mN}$, roughly the weight of an object of $30\,\text{mg}$, for example a grain of rice. It is small, but in a vacuum it acts continuously, so a light probe accelerates to a high speed over months. If the sail absorbed the light, the force would be half, $145\,\mu\text{N}$.
Check your understanding
Exit ticket
Answer on your own first and only then open the answer.
- Name one phenomenon in which light behaves like a wave, and one in which it behaves like a particle.
Answer
Like a wave: the fringe pattern of the double-slit experiment, that is, interference (also diffraction). Like a particle: the photoelectric effect, the Compton effect or the single hits of weak light on a screen.
- Which photon has the greater momentum: that of red light ($700\,\text{nm}$) or that of blue light ($450\,\text{nm}$)? Justify your answer.
Answer
That of blue light. The momentum of a photon is $p = \tfrac{h}{\lambda}$, so a shorter wavelength gives a greater momentum: the momentum of the blue photon is $700 : 450 \approx 1.6$ times as large.
1. Concept
2. Compton effect
An X-ray photon of wavelength $0.0600\,\text{nm}$ scatters from an almost free electron. The wavelength of the scattered photon is $0.0624\,\text{nm}$.
How much kinetic energy does the electron receive? Give the answer in electronvolts.
Hint 1: a guiding question
Where does the energy lost by the photon go?
Hint 2: what is the connection
when the wavelengths are in metres.
Hint 3: an intermediate step
The energy of the incident photon is $3.3108 \cdot 10^{-15}\,\text{J}$ and that of the scattered one $3.1835 \cdot 10^{-15}\,\text{J}$. Divide the difference by $1.602 \cdot 10^{-19}\,\text{J}$.
Show the solution
Given
the wavelength of the incident X-ray photon $\lambda = 0.0600\,\text{nm}$, of the scattered one $\lambda' = 0.0624\,\text{nm}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^8\,\text{m/s}$, $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$
Asked
the kinetic energy $E_\text{k}$ received by the electron in electronvolts
Principle
In the Compton effect, the photon collides with an almost free electron. Energy is conserved, so the electron receives the energy that the photon loses: $hf = hf' + E_\text{k}$. The energy of each photon is obtained from its wavelength.
Formula
from which
Conversions
$0.0600\,\text{nm} = 0.0600 \cdot 10^{-9}\,\text{m}$ and $0.0624\,\text{nm} = 0.0624 \cdot 10^{-9}\,\text{m}$. Finally joules to electronvolts: $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$.
Substitution
In electronvolts:
Answer
$7.9 \cdot 10^2\,\text{eV}$
Two significant figures, because the difference of the wavelengths $0.0024\,\text{nm}$ is known only to two figures.
Reasonableness
The energy of the incident photon is
and the wavelength grows by $4\,\%$, so about $4\,\%$ of the energy, about $0.8\,\text{keV}$, is transferred to the electron. The electron gains energy and the photon loses it, so $E_\text{k}$ is positive.
Suggestion: one answer does not show everything you can do, but review the theory summary and do the core exercises.
Optional 3. Evaluate the claim
Your friend claims: “Light is really a wave. The photon is just a calculation aid needed in the photoelectric effect.” Evaluate the claim using the double-slit experiment and the Compton effect.
Model answer
The claim is only partly correct. The wave model explains the fringe pattern of the double-slit experiment, but in an experiment with weak light the light is detected as single photons that hit the screen one at a time at one place. In the Compton effect, the wavelength of the scattered X-rays grows with the scattering angle, which the wave model does not explain. In the photon model the photon gives the electron part of its energy and its momentum $p = \tfrac{h}{\lambda}$, as in a collision. So the photon is not merely a calculation aid. Light is neither only a wave nor only bullets: the wave gives the probability, and the photon is a quantum that is detected at one place. This is wave-particle duality.
Assess your own answer
Tick the points your own answer already contains.
Review the theory summary and practise with the exercises.
Exercises
Level 1 · basic practice 5 exercises
One formula or one step of reasoning. Core exercises first, the rest is extra practice.
- Exercise 4-1Core
A green laser pointer emits light of wavelength $532\,\text{nm}$. What is the momentum of one photon? Give the answer in units of $10^{-27}\,\text{kg}\cdot\text{m/s}$.
Answer
$1.25 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$
Show the solution
Given
wavelength $\lambda = 532\,\text{nm}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$
Asked
the momentum $p$ of a photon in units of $10^{-27}\,\text{kg}\cdot\text{m/s}$
Principle
A photon has no mass, but it has a momentum that depends only on the wavelength. The light has one wavelength, so every photon has the same momentum.
Formula
$p = \dfrac{h}{\lambda}$
Conversions
$532\,\text{nm} = 532 \cdot 10^{-9}\,\text{m}$. Unit: $\tfrac{\text{J}\,\text{s}}{\text{m}} = \text{kg}\cdot\text{m/s}$.
Substitution
$p$$= \dfrac{6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}{532 \cdot 10^{-9}\,\text{m}}$$= 1.2455 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$Answer
$1.25 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$
Three significant figures as in the wavelength $532\,\text{nm}$.
Reasonableness
According to the theory section, the momentum of a photon of $500\,\text{nm}$ is $1.3 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$. The wavelength $532\,\text{nm}$ is a little longer, so the momentum is a little smaller, as it should be. The value is very small because $h$ is very small.
- Exercise 4-2Core
The molybdenum anode of an X-ray tube strongly emits X-rays of wavelength $0.0711\,\text{nm}$.
What is the momentum of a photon of this radiation? Give the answer in units of $10^{-24}\,\text{kg}\cdot\text{m/s}$.
Answer
$9.32 \cdot 10^{-24}\,\text{kg}\cdot\text{m/s}$
Show the solution
Given
wavelength of the X-rays $\lambda = 0.0711\,\text{nm}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$
Asked
the momentum $p$ of a photon in units of $10^{-24}\,\text{kg}\cdot\text{m/s}$
Principle
The momentum of a photon depends only on its wavelength, so the same formula applies to X-rays as to visible light.
Formula
$p = \dfrac{h}{\lambda}$
Conversions
$0.0711\,\text{nm} = 0.0711 \cdot 10^{-9}\,\text{m}$
Substitution
$p$$= \dfrac{6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}{0.0711 \cdot 10^{-9}\,\text{m}}$$= 9.3193 \cdot 10^{-24}\,\text{kg}\cdot\text{m/s}$Answer
$9.32 \cdot 10^{-24}\,\text{kg}\cdot\text{m/s}$
Three significant figures as in the wavelength $0.0711\,\text{nm}$.
Reasonableness
The wavelength is about $7000$ times shorter than that of visible light, $500\,\text{nm}$, so the momentum is about $7000$ times greater:
$7000 \cdot 1.3 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$$\approx 9 \cdot 10^{-24}\,\text{kg}\cdot\text{m/s}$ - Exercise 4-3
The momentum of a photon is $3.60 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$. What is the wavelength of the photon? Give the answer in nanometres.
Answer
$184\,\text{nm}$
Show the solution
Given
momentum of the photon $p = 3.60 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$
Asked
the wavelength $\lambda$ of the photon in nanometres
Principle
The momentum and the wavelength of a photon are inversely proportional, so the wavelength follows from the known momentum.
Formula
$p = \dfrac{h}{\lambda}$, from which $\lambda = \dfrac{h}{p}$
Conversions
$\tfrac{\text{J}\,\text{s}}{\text{kg}\cdot\text{m/s}}$$= \tfrac{\text{kg}\cdot\text{m}^2/\text{s}}{\text{kg}\cdot\text{m/s}}$$= \text{m}$Finally $1\,\text{m} = 10^9\,\text{nm}$.
Substitution
$\lambda$$= \dfrac{6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}{3.60 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}}$$= 1.8406 \cdot 10^{-7}\,\text{m}$$= 184.06\,\text{nm}$Answer
$184\,\text{nm}$
Three significant figures as in the momentum $3.60 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$.
Reasonableness
The momentum is about three times that of a photon of visible light of $500\,\text{nm}$ ($1.3 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$), so the wavelength must be about a third: $500\,\text{nm} : 2.7 \approx 185\,\text{nm}$. The photon belongs to ultraviolet radiation.
- Exercise 4-4
The momentum of a photon is $2.00 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$. What is the energy of the photon? Give the answer in units of $10^{-19}\,\text{J}$.
Answer
$6.00 \cdot 10^{-19}\,\text{J}$
Show the solution
Given
momentum of the photon $p = 2.00 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$; $c = 2.998 \cdot 10^8\,\text{m/s}$
Asked
the energy $E$ of the photon in units of $10^{-19}\,\text{J}$
Principle
The momentum of a photon is its energy divided by the speed of light, because a photon always travels at the speed $c$ and has no mass.
Formula
$p = \dfrac{E}{c}$, from which $E = pc$
Conversions
None needed: the quantities are in SI units, and
$\text{kg}\cdot\tfrac{\text{m}}{\text{s}} \cdot \tfrac{\text{m}}{\text{s}}$$= \text{kg}\cdot\tfrac{\text{m}^2}{\text{s}^2}$$= \text{J}$Substitution
$E$$= 2.00 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s} \cdot 2.998 \cdot 10^8\,\text{m/s}$$= 5.996 \cdot 10^{-19}\,\text{J}$Answer
$6.00 \cdot 10^{-19}\,\text{J}$
Three significant figures as in the momentum $2.00 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$.
Reasonableness
In electronvolts the energy is
$\tfrac{5.996 \cdot 10^{-19}\,\text{J}}{1.602 \cdot 10^{-19}\,\text{J}} \cdot 1\,\text{eV}$$\approx 3.7\,\text{eV}$The wavelength of the photon is
$\lambda$$= \tfrac{h}{p}$$= \tfrac{6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}{2.00 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}}$$\approx 331\,\text{nm}$so the photon is in the near ultraviolet, a little beyond the violet end of visible light ($400\,\text{nm}$). This is reasonable, because the momentum is about one and a half times that of a photon of $500\,\text{nm}$.
- Exercise 4-5
A laser of power $1.0\,\text{mW}$ emits photons, each of energy $3.14 \cdot 10^{-19}\,\text{J}$. How many photons does the laser emit per second? Give the answer in units of $10^{15}\,\tfrac{1}{\text{s}}$.
Answer
$3.2 \cdot 10^{15}\,\tfrac{1}{\text{s}}$
Show the solution
Given
power of the laser $P = 1.0\,\text{mW}$, energy of one photon $hf = 3.14 \cdot 10^{-19}\,\text{J}$
Asked
the number of photons per second $\tfrac{N}{t}$ in units of $10^{15}\,\tfrac{1}{\text{s}}$
Principle
The energy of light travels as photons. The power is the total energy of the photons emitted per second, so the number of photons per second is the power divided by the energy of one photon.
Formula
$P = \dfrac{N}{t} \cdot hf$, from which $\dfrac{N}{t} = \dfrac{P}{hf}$
Conversions
$1.0\,\text{mW}$$= 1.0 \cdot 10^{-3}\,\text{W}$$= 1.0 \cdot 10^{-3}\,\tfrac{\text{J}}{\text{s}}$Substitution
$\dfrac{N}{t}$$= \dfrac{1.0 \cdot 10^{-3}\,\text{J/s}}{3.14 \cdot 10^{-19}\,\text{J}}$$= 3.185 \cdot 10^{15}\,\tfrac{1}{\text{s}}$Answer
$3.2 \cdot 10^{15}\,\tfrac{1}{\text{s}}$
Two significant figures as in the power $1.0\,\text{mW}$.
Reasonableness
The photon energy $3.14 \cdot 10^{-19}\,\text{J}$ is that of a photon of red light. One millijoule therefore takes thousands of trillions of photons, so ordinary light looks continuous. The number is of the same order of magnitude as the number of photons of a laser pointer mentioned in the example of the chapter.
Level 2 · applied practice 4 exercises
Two steps, working with measured data, a unit conversion or rearranging a formula. Core exercises first, the rest is extra practice.
- Exercise 4-6Core
An X-ray photon of wavelength $0.0711\,\text{nm}$ scatters from an almost free electron in graphite. The wavelength of the scattered photon is $0.0735\,\text{nm}$.
How much kinetic energy does the electron receive? Give the answer in electronvolts.
Answer
$5.7 \cdot 10^2\,\text{eV}$
Show the solution
Given
wavelength of the incident photon $\lambda = 0.0711\,\text{nm}$, of the scattered one $\lambda' = 0.0735\,\text{nm}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^8\,\text{m/s}$, $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$
Asked
the kinetic energy $E_\text{k}$ received by the electron in electronvolts
Principle
In the Compton effect, the photon collides with an almost free electron. Energy is conserved, so the electron receives the energy that the photon loses: $hf = hf' + E_\text{k}$. The energies of the photons are obtained from the wavelengths.
Formula
$\dfrac{hc}{\lambda}$$= \dfrac{hc}{\lambda'} + E_\text{k}$from which
$E_\text{k}$$= \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda'}$Conversions
$0.0711\,\text{nm} = 0.0711 \cdot 10^{-9}\,\text{m}$ and $0.0735\,\text{nm} = 0.0735 \cdot 10^{-9}\,\text{m}$. Finally joules to electronvolts by substituting $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$.
Substitution
$hc$$= 6.626 \cdot 10^{-34}\,\text{J}\,\text{s} \cdot 2.998 \cdot 10^8\,\text{m/s}$$= 1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}$so
$E_\text{k}$$= \dfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{0.0711 \cdot 10^{-9}\,\text{m}} - \dfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{0.0735 \cdot 10^{-9}\,\text{m}}$$= 2.7939 \cdot 10^{-15}\,\text{J} - 2.7027 \cdot 10^{-15}\,\text{J}$$= 9.123 \cdot 10^{-17}\,\text{J}$and in electronvolts
$E_\text{k}$$= \dfrac{9.123 \cdot 10^{-17}\,\text{J}}{1.602 \cdot 10^{-19}\,\text{J}} \cdot 1\,\text{eV}$$= 569.5\,\text{eV}$Answer
$5.7 \cdot 10^2\,\text{eV}$
Two significant figures, because the difference of the wavelengths $0.0024\,\text{nm}$ is known only to two figures.
Reasonableness
The energy of the incident photon is
$\tfrac{2.7939 \cdot 10^{-15}\,\text{J}}{1.602 \cdot 10^{-19}\,\text{J}} \cdot 1\,\text{eV}$$\approx 17.4\,\text{keV}$The wavelength grows by about $3\,\%$, so the photon loses about $3\,\%$ of its energy, about $0.6\,\text{keV}$. The electron gains energy, so the result is positive.
- Exercise 4-7Core
The decay of caesium-137 produces a gamma photon of energy $662\,\text{keV}$. What is the momentum of the photon? Give the answer in units of $10^{-22}\,\text{kg}\cdot\text{m/s}$.
Answer
$3.54 \cdot 10^{-22}\,\text{kg}\cdot\text{m/s}$
Show the solution
Given
energy of the gamma photon $E = 662\,\text{keV}$; $c = 2.998 \cdot 10^8\,\text{m/s}$, $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$
Asked
the momentum $p$ of the photon in units of $10^{-22}\,\text{kg}\cdot\text{m/s}$
Principle
The momentum of a photon is its energy divided by the speed of light. The energy is given in electronvolts, so it is first converted to joules.
Formula
$p = \dfrac{E}{c}$
Conversions
$662\,\text{keV}$$= 662 \cdot 10^3\,\text{eV}$$= 662 \cdot 10^3 \cdot 1.602 \cdot 10^{-19}\,\text{J}$$= 1.0605 \cdot 10^{-13}\,\text{J}$Substitution
$p$$= \dfrac{1.0605 \cdot 10^{-13}\,\text{J}}{2.998 \cdot 10^8\,\text{m/s}}$$= 3.5374 \cdot 10^{-22}\,\text{kg}\cdot\text{m/s}$Answer
$3.54 \cdot 10^{-22}\,\text{kg}\cdot\text{m/s}$
Three significant figures as in the energy $662\,\text{keV}$.
Reasonableness
The energy of the gamma photon is about $40$ times the energy $17.4\,\text{keV}$ of the X-ray photon, so the momentum is about $40$ times the momentum of the X-ray photon of level 1 task 2:
$40 \cdot 9.3 \cdot 10^{-24}\,\text{kg}\cdot\text{m/s}$$\approx 3.7 \cdot 10^{-22}\,\text{kg}\cdot\text{m/s}$ - Exercise 4-8
The wavelength of infrared light travelling in an optical fibre is $1550\,\text{nm}$. A photon hits a mirror perpendicularly and is reflected straight back.
How much momentum does the mirror receive from one photon? Give the answer in units of $10^{-28}\,\text{kg}\cdot\text{m/s}$.
Answer
$8.55 \cdot 10^{-28}\,\text{kg}\cdot\text{m/s}$
Show the solution
Given
wavelength $\lambda = 1550\,\text{nm}$, the photon is reflected perpendicularly back; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$
Asked
the momentum $\Delta p$ received by the mirror from one photon in units of $10^{-28}\,\text{kg}\cdot\text{m/s}$
Principle
Momentum is conserved. In reflection the momentum of the photon reverses, so its change is $p - (-p) = 2p$. The mirror receives a momentum of the same magnitude in the direction of arrival of the photon, because the total momentum is conserved. The wavelength of the reflected photon is the same, because a heavy mirror hardly receives any energy.
Formula
$\Delta p = p - (-p)$, where $p = \dfrac{h}{\lambda}$, so $\Delta p = 2 \cdot \dfrac{h}{\lambda}$
Conversions
$1550\,\text{nm} = 1550 \cdot 10^{-9}\,\text{m}$
Substitution
$\Delta p$$= \dfrac{2 \cdot 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}{1550 \cdot 10^{-9}\,\text{m}}$$= 8.550 \cdot 10^{-28}\,\text{kg}\cdot\text{m/s}$Answer
$8.55 \cdot 10^{-28}\,\text{kg}\cdot\text{m/s}$
Three significant figures as in the wavelength $1550\,\text{nm}$ (the constant $2$ is exact).
Reasonableness
The wavelength is about three times that of green light, $532\,\text{nm}$, so the momentum of one photon is about a third of $1.25 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$, that is, about $4 \cdot 10^{-28}\,\text{kg}\cdot\text{m/s}$, and the reflection doubles it.
- Exercise 4-9
A sensitive light detector counts $2.40 \cdot 10^{4}$ photons per second. The wavelength of the light is $550\,\text{nm}$.
What is the power of the light arriving at the detector? Give the answer in units of $10^{-15}\,\text{W}$.
Answer
$8.67 \cdot 10^{-15}\,\text{W}$
Show the solution
Given
photons per second $\tfrac{N}{t} = 2.40 \cdot 10^4\,\tfrac{1}{\text{s}}$, wavelength $\lambda = 550\,\text{nm}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^8\,\text{m/s}$
Asked
the power $P$ of the light in units of $10^{-15}\,\text{W}$
Principle
The power is the total energy of the photons arriving per second. The energy of each photon is $hf = \tfrac{hc}{\lambda}$, because the light has one wavelength.
Formula
$P = \dfrac{N}{t} \cdot hf$, where $hf = \dfrac{hc}{\lambda}$, so
$P$$= \dfrac{N}{t} \cdot \dfrac{hc}{\lambda}$Conversions
$550\,\text{nm} = 550 \cdot 10^{-9}\,\text{m}$
Substitution
$hc$$= 6.626 \cdot 10^{-34}\,\text{J}\,\text{s} \cdot 2.998 \cdot 10^8\,\text{m/s}$$= 1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}$so
$hf$$= \dfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{550 \cdot 10^{-9}\,\text{m}}$$= 3.6118 \cdot 10^{-19}\,\text{J}$$P$$= 2.40 \cdot 10^4\,\tfrac{1}{\text{s}} \cdot 3.6118 \cdot 10^{-19}\,\text{J}$$= 8.668 \cdot 10^{-15}\,\text{W}$Answer
$8.67 \cdot 10^{-15}\,\text{W}$
Three significant figures as in the number of photons $2.40 \cdot 10^4$ and the wavelength $550\,\text{nm}$.
Reasonableness
The power is a millionth of a billionth of a watt, so such light is too weak to be seen by the eye. A check the other way round:
$\tfrac{8.668 \cdot 10^{-15}\,\text{W}}{3.6118 \cdot 10^{-19}\,\text{J}}$$= 2.40 \cdot 10^4\,\tfrac{1}{\text{s}}$
Level 3 · challenging 3 exercises
Optional: application, two ideas together or a new situation where you reason for yourself.
- Exercise 4-10
In laser cooling, laser light of wavelength $589\,\text{nm}$ is directed against a sodium atom. The atom absorbs one photon at a time, and each photon reduces the momentum of the atom by the momentum of the photon.
The mass of the sodium atom is $3.82 \cdot 10^{-26}\,\text{kg}$, and its initial speed is $3.00 \cdot 10^{2}\,\text{m/s}$.
How many photons must the atom absorb to stop? Assume that all the photons come straight against it, and ignore the effect of the photons re-emitted by the atom, because they leave in random directions. Give the answer in units of $10^{4}$.
Answer
$1.02 \cdot 10^4$
Show the solution
Given
wavelength $\lambda = 589\,\text{nm}$, mass of the atom $m = 3.82 \cdot 10^{-26}\,\text{kg}$, initial speed $v_0 = 3.00 \cdot 10^2\,\text{m/s}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$
Asked
the number $N$ of photons needed, in units of $10^4$
Principle
Momentum is conserved. When the atom absorbs a photon coming against it, the momentum of the atom decreases by the momentum $\tfrac{h}{\lambda}$ of the photon, so its speed decreases by $\Delta v = \tfrac{h}{\lambda m}$. The atom stops when $N$ photons have reduced its momentum $m v_0$ to zero. The re-emitted photons leave in random directions, so on average their effects cancel.
Formula
$m v_0 = N p$, where $p = \dfrac{h}{\lambda}$, so $N = \dfrac{m v_0}{p}$
Conversions
$589\,\text{nm} = 589 \cdot 10^{-9}\,\text{m}$
Substitution
$p$$= \dfrac{6.626 \cdot 10^{-34}\,\text{J}\,\text{s}}{589 \cdot 10^{-9}\,\text{m}}$$= 1.1250 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$$m v_0$$= 3.82 \cdot 10^{-26}\,\text{kg} \cdot 3.00 \cdot 10^2\,\text{m/s}$$= 1.146 \cdot 10^{-23}\,\text{kg}\cdot\text{m/s}$so
$N$$= \dfrac{1.146 \cdot 10^{-23}\,\text{kg}\cdot\text{m/s}}{1.1250 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}}$$= 1.0187 \cdot 10^4$Answer
$1.02 \cdot 10^4$
Photons, three significant figures as in the mass, the speed and the wavelength.
Reasonableness
One photon changes the speed of the atom by
$\tfrac{1.1250 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}}{3.82 \cdot 10^{-26}\,\text{kg}}$$\approx 0.029\,\text{m/s}$so stopping $300\,\text{m/s}$ takes about $300 : 0.029 \approx 10\,000$ photons. An atom can absorb and emit photons millions of times per second, so the braking takes only about a millisecond.
- Exercise 4-11
An X-ray photon of wavelength $0.0500\,\text{nm}$ scatters from an almost free electron. The electron receives the kinetic energy $1.00\,\text{keV}$.
What is the wavelength of the scattered photon? Give the answer in nanometres to four decimal places.
Answer
$0.0521\,\text{nm}$
Show the solution
Given
wavelength of the incident photon $\lambda = 0.0500\,\text{nm}$, kinetic energy of the electron $E_\text{k} = 1.00\,\text{keV}$; $h = 6.626 \cdot 10^{-34}\,\text{J}\,\text{s}$, $c = 2.998 \cdot 10^8\,\text{m/s}$, $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$
Asked
the wavelength $\lambda'$ of the scattered photon in nanometres
Principle
In the Compton effect, energy is conserved: the energy of the incident photon is shared between the scattered photon and the electron. The wavelength of the scattered photon follows from its energy.
Formula
$\dfrac{hc}{\lambda}$$= \dfrac{hc}{\lambda'} + E_\text{k}$from which
$\lambda'$$= \dfrac{hc}{hc/\lambda - E_\text{k}}$Conversions
$0.0500\,\text{nm}$$= 0.0500 \cdot 10^{-9}\,\text{m}$$1.00\,\text{keV}$$= 1.00 \cdot 10^3 \cdot 1.602 \cdot 10^{-19}\,\text{J}$$= 1.602 \cdot 10^{-16}\,\text{J}$Substitution
$\dfrac{hc}{\lambda}$$= \dfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{0.0500 \cdot 10^{-9}\,\text{m}}$$= 3.9729 \cdot 10^{-15}\,\text{J}$so the energy of the scattered photon is
$3.9729 \cdot 10^{-15}\,\text{J} - 0.1602 \cdot 10^{-15}\,\text{J}$$= 3.8127 \cdot 10^{-15}\,\text{J}$$\lambda'$$= \dfrac{1.9865 \cdot 10^{-25}\,\text{J}\,\text{m}}{3.8127 \cdot 10^{-15}\,\text{J}}$$= 5.210 \cdot 10^{-11}\,\text{m}$$= 0.05210\,\text{nm}$Answer
$0.0521\,\text{nm}$
Three significant figures as in the wavelength $0.0500\,\text{nm}$ and the energy $1.00\,\text{keV}$.
Reasonableness
The energy of the incident photon is about $24.8\,\text{keV}$, and the electron receives $1.00\,\text{keV}$ of it, about $4\,\%$. The wavelength of the photon therefore grows by about $4\,\%$: $0.0500\,\text{nm} \cdot 1.04 = 0.0520\,\text{nm}$. The wavelength grows, as it must in the Compton effect.
- Exercise 4-12
A hydrogen atom at rest moves from the energy level $n = 3$ to the level $n = 2$ and emits a photon. The energy levels of hydrogen are $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$. The atom receives a momentum equal in magnitude to that of the photon and in the opposite direction, that is, a recoil.
What is the speed of the atom after the photon has left? The mass of a hydrogen atom is $1.67 \cdot 10^{-27}\,\text{kg}$. Give the answer in metres per second.
Answer
$0.604\,\text{m/s}$
Show the solution
Given
transition $n = 3 \to n = 2$, $E_n = -\tfrac{13.6\,\text{eV}}{n^2}$, mass of the hydrogen atom $m = 1.67 \cdot 10^{-27}\,\text{kg}$, the atom initially at rest; $c = 2.998 \cdot 10^8\,\text{m/s}$, $1\,\text{eV} = 1.602 \cdot 10^{-19}\,\text{J}$
Asked
the recoil speed $v$ of the atom in metres per second
Principle
The energy of the photon is the difference of the energy levels. Momentum is conserved: the total momentum is initially zero, so after the photon has left, the momentum $mv$ of the atom is equal in magnitude and opposite in direction to the momentum $\tfrac{E}{c}$ of the photon. The kinetic energy received by the atom is so small that the energy of the photon is in practice the difference of the levels.
Formula
$E = E_3 - E_2$ and $mv = \dfrac{E}{c}$, from which $v = \dfrac{E}{mc}$
Conversions
$E_3$$= -\tfrac{13.6\,\text{eV}}{9}$$= -1.511\,\text{eV}$$E_2$$= -\tfrac{13.6\,\text{eV}}{4}$$= -3.400\,\text{eV}$so
$E$$= 1.889\,\text{eV}$$= 1.889 \cdot 1.602 \cdot 10^{-19}\,\text{J}$$= 3.026 \cdot 10^{-19}\,\text{J}$Substitution
$v$$= \dfrac{3.026 \cdot 10^{-19}\,\text{J}}{1.67 \cdot 10^{-27}\,\text{kg} \cdot 2.998 \cdot 10^8\,\text{m/s}}$$= 0.6044\,\text{m/s}$Answer
$0.604\,\text{m/s}$
Three significant figures as in the mass $1.67 \cdot 10^{-27}\,\text{kg}$ and the energy $13.6\,\text{eV}$.
Reasonableness
The momentum of the photon is
$\tfrac{3.026 \cdot 10^{-19}\,\text{J}}{2.998 \cdot 10^8\,\text{m/s}}$$\approx 1.0 \cdot 10^{-27}\,\text{kg}\cdot\text{m/s}$roughly equal to the mass of a hydrogen atom in kilograms, so the speed is less than $1\,\text{m/s}$. It is very small compared with the speeds of the thermal motion of atoms, which at room temperature are kilometres per second, so the recoil does not show in everyday observations. The photon is red light, $656\,\text{nm}$.
Photographs
- Electron microscope at Oak Ridge National Lab. (7209791480). Oak Ridge National Laboratory (U.S. Department of Energy). source, public domain
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