# 29.4 Photon momentum  (Page 3/5)

 Page 3 / 5

## Relativistic photon momentum

There is a relationship between photon momentum $p$ and photon energy $E$ that is consistent with the relation given previously for the relativistic total energy of a particle as ${E}^{2}=\left(\text{pc}{\right)}^{2}+\left(\text{mc}{\right)}^{2}$ . We know $m$ is zero for a photon, but $p$ is not, so that ${E}^{2}=\left(\text{pc}{\right)}^{2}+\left(\text{mc}{\right)}^{2}$ becomes

$E=\text{pc},$

or

$p=\frac{E}{c}\left(photons\right).$

To check the validity of this relation, note that $E=\text{hc}/\lambda$ for a photon. Substituting this into $p=E/c$ yields

$p=\left(\text{hc}/\lambda \right)/c=\frac{h}{\lambda },$

as determined experimentally and discussed above. Thus, $p=E/c$ is equivalent to Compton’s result $p=h/\lambda$ . For a further verification of the relationship between photon energy and momentum, see [link] .

## Photon detectors

Almost all detection systems talked about thus far—eyes, photographic plates, photomultiplier tubes in microscopes, and CCD cameras—rely on particle-like properties of photons interacting with a sensitive area. A change is caused and either the change is cascaded or zillions of points are recorded to form an image we detect. These detectors are used in biomedical imaging systems, and there is ongoing research into improving the efficiency of receiving photons, particularly by cooling detection systems and reducing thermal effects.

## Photon energy and momentum

Show that $p=E/c$ for the photon considered in the [link] .

Strategy

We will take the energy $E$ found in [link] , divide it by the speed of light, and see if the same momentum is obtained as before.

Solution

Given that the energy of the photon is 2.48 eV and converting this to joules, we get

$p=\frac{E}{c}=\frac{\left(2.48 eV\right)\left(1\text{.}\text{60}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J/eV}\right)}{3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}=\text{1}\text{.}\text{33}×{\text{10}}^{\text{–27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}.$

Discussion

This value for momentum is the same as found before (note that unrounded values are used in all calculations to avoid even small rounding errors), an expected verification of the relationship $p=E/c$ . This also means the relationship between energy, momentum, and mass given by ${E}^{2}=\left(\text{pc}{\right)}^{2}+\left(\text{mc}{\right)}^{2}$ applies to both matter and photons. Once again, note that $p$ is not zero, even when $m$ is.

## Problem-solving suggestion

Note that the forms of the constants $h=\text{4}\text{.}\text{14}×{\text{10}}^{\text{–15}}\phantom{\rule{0.25em}{0ex}}\text{eV}\cdot \text{s}$ and $\text{hc}=\text{1240 eV}\cdot \text{nm}$ may be particularly useful for this section’s Problems and Exercises.

## Section summary

• Photons have momentum, given by $p=\frac{h}{\lambda }$ , where $\lambda$ is the photon wavelength.
• Photon energy and momentum are related by $p=\frac{E}{c}$ , where $E=\text{hf}=\text{hc}/\lambda$ for a photon.

## Conceptual questions

Which formula may be used for the momentum of all particles, with or without mass?

Is there any measurable difference between the momentum of a photon and the momentum of matter?

Why don’t we feel the momentum of sunlight when we are on the beach?

## Problems&Exercises

(a) Find the momentum of a 4.00-cm-wavelength microwave photon. (b) Discuss why you expect the answer to (a) to be very small.

(a) $\text{1.66}×{\text{10}}^{-\text{32}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$

(b) The wavelength of microwave photons is large, so the momentum they carry is very small.

(a) What is the momentum of a 0.0100-nm-wavelength photon that could detect details of an atom? (b) What is its energy in MeV?

(a) What is the wavelength of a photon that has a momentum of $5\text{.}\text{00}×{\text{10}}^{-\text{29}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$ ? (b) Find its energy in eV.

(a) 13.3 μm

(b) $9\text{.}\text{38}×{\text{10}}^{-2}$ eV

(a) A $\gamma$ -ray photon has a momentum of $8\text{.}\text{00}×{\text{10}}^{-\text{21}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$ . What is its wavelength? (b) Calculate its energy in MeV.

(a) Calculate the momentum of a photon having a wavelength of $2\text{.}\text{50 μm}$ . (b) Find the velocity of an electron having the same momentum. (c) What is the kinetic energy of the electron, and how does it compare with that of the photon?

(a) $2\text{.}\text{65}×{\text{10}}^{-\text{28}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}$

(b) 291 m/s

(c) electron $3\text{.}\text{86}×{\text{10}}^{-\text{26}}\phantom{\rule{0.25em}{0ex}}\text{J}$ , photon $7\text{.}\text{96}×{\text{10}}^{-\text{20}}\phantom{\rule{0.25em}{0ex}}\text{J}$ , ratio $2\text{.}\text{06}×{\text{10}}^{6}$

Repeat the previous problem for a 10.0-nm-wavelength photon.

(a) Calculate the wavelength of a photon that has the same momentum as a proton moving at 1.00% of the speed of light. (b) What is the energy of the photon in MeV? (c) What is the kinetic energy of the proton in MeV?

(a) $1\text{.}\text{32}×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{m}$

(b) 9.39 MeV

(c) $4.70×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{MeV}$

(a) Find the momentum of a 100-keV x-ray photon. (b) Find the equivalent velocity of a neutron with the same momentum. (c) What is the neutron’s kinetic energy in keV?

Take the ratio of relativistic rest energy, $E={\mathrm{\gamma mc}}^{2}$ , to relativistic momentum, $p=\gamma \text{mu}$ , and show that in the limit that mass approaches zero, you find $E/p=c$ .

$E={\mathrm{\gamma mc}}^{2}$ and $P=\mathrm{\gamma mu}$ , so

$\frac{E}{P}=\frac{{\text{γmc}}^{2}}{\text{γmu}}=\frac{{\text{c}}^{2}}{\text{u}}.$

As the mass of particle approaches zero, its velocity $u$ will approach $c$ , so that the ratio of energy to momentum in this limit is

${lim}_{m\to 0}\frac{E}{P}=\frac{{c}^{2}}{c}=c$

which is consistent with the equation for photon energy.

Consider a space sail such as mentioned in [link] . Construct a problem in which you calculate the light pressure on the sail in ${\text{N/m}}^{2}$ produced by reflecting sunlight. Also calculate the force that could be produced and how much effect that would have on a spacecraft. Among the things to be considered are the intensity of sunlight, its average wavelength, the number of photons per square meter this implies, the area of the space sail, and the mass of the system being accelerated.

Unreasonable Results

A car feels a small force due to the light it sends out from its headlights, equal to the momentum of the light divided by the time in which it is emitted. (a) Calculate the power of each headlight, if they exert a total force of $2\text{.}\text{00}×{\text{10}}^{-2}\phantom{\rule{0.25em}{0ex}}\text{N}$ backward on the car. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

(a) $3\text{.}\text{00}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{W}$

(b) Headlights are way too bright.

(c) Force is too large.

Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.
hour glass, pendulum clock, atomic clock?
S.M
tnks
David
how did they solve for "t" after getting 67.6=.5(Voy + 0)t
Find the following for path D in [link] : (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
the topic is kinematics
David
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Lohitha
just check the chpt. 13 kinetic theory of matter it's there
David
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no it is derived
Abdul
no
Nisha
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David
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Emmanuel
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David
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Longwar
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how about a conceptual framework can you simplify for me? needed please
Villaflor
Hello what happens when electrone stops its rotation around its nucleus if it possible how
Afzal
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Villaflor
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S.M
not possible to fix electron position in space,
S.M
Physics
Beatriz
yes of course Villa flor
David
equations of kinematics for constant acceleration
A bottle full of water weighs 45g when full of mercury,it weighs 360g.if the empty bottle weighs 20g.calculate the relative density of mercury and the density of mercury....pls I need help
well You know the density of water is 1000kg/m^3.And formula for density is density=mass/volume Then we must calculate volume of bottle and mass of mercury: Volume of bottle is (45-20)/1000000=1/40000 mass of mercury is:(360-20)/1000 kg density of mercury:(340/1000):1/50000=(340•40000):1000=13600
Sobirjon
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Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture...take density of water as 1g/cm3 and density of liquid 1.2g/cm3
Lila
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who can help me with my problem about acceleration?
ok
Nicholas
how to solve this... a car is heading north then smoothly made a westward turn during the travel the speed of the car remains constant at 1.5km/h what is the acceleration of the car? the total travel time of the car as it smoothly changed its direction is 15 minutes
Vann
i think the acceleration is 0 since the car does not change its speed unless there are other conditions
Ben
yes I have to agree, the key phrase is, "the speed of the car remains constant...," all other information is not needed to conclude that acceleration remains at 0 during the entire time
Luis
who can help me with a relative density question
Lila
1cm3 sample of tin lead alloy has mass 8.5g.the relative density of tin is 7.3 and that of lead is 11.3.calculate the percentage by weight of tin in the alloy. assuming that there is no change of volume when the metals formed the alloy
Lila
morning, what will happen to the volume of an ice block when heat is added from -200°c to 0°c... Will it volume increase or decrease?
no
Emmanuel
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Kate
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Kate
No
Emmanuel
I think it is neither decreases nor increases ,it remains in the same volume because of its crystal structure
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture. take density of water as 1g/cm3 and density of liquid as 1.2g/cm3
Lila
Sorry what does it means"no changes in volume occured"?
Sobirjon
volume can be the amount of space occupied by an object. But when an object does not change in shape it will still occupy the same space. Thats why the volume will still remain the same
Ben
Most soilds expand when heated but if it changes state at 0C it will have less volume. Ice floats because it is less dense ie a larger mass per unit volume.
Richard
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v=d/t
Emeka
Villaflor
Villaflor
v=d/t
Nisha
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H.C
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ABDUL
If someone has not studied Mathematics enough yet, should theu study it first then study Phusics or Study Basics of Physics whilst srudying Math as well?
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Nuru
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Thomas
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Thomas
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Anand
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Nuru
A hydrometer of mass 0.15kg and uniform cross sectional area of 0.0025m2 displaced in water of density 1000kg/m3.what depth will the hydrometer sink
Lila
16.66 meters?
Darshik
16.71m2
aways
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aways
the mass is stretched a distance of 8cm and held what is the potential energy? quick answer
aways
oscillation is a to and fro movement, it can also be referred to as vibration. e.g loaded string, loaded test tube or an hinged door