Modern Physics, Chapter Notes, Class 12, Physics (IIT-JEE & AIPMT) PDF Download

Nature of light

It was a matter of great interest for scientists of know that what exactly from the light is made up of or how the light behaves. This is briefly described over here

Newton's Corpuscular theory

Newton was the first scientist who said that light is made up of tiny elastic particles called "Corpuscles" which travels with the velocity of light. So according to Newtons, light is a particle.

Huygen's wave theory

Huygen was a scientist working parallel to Newton who come with a drastically different idea for nature of light & said that light is not a particle but a wave.

Maxwell's electromagnetic wave theory

During the time of Hygen, his views regarding nature of light were not accepted as newton was a popular scientist of his time. but, when maxwell asserted that light is a electromagnetic wave, scientists started believing that light is a wave.

Max Planck's quantum theory of light

Once again when scientists started believing that the light is a wave max plank came with different idea & asserted that light is not a wave but a photon (i.e. a particle) which he proved through balck body radiation spectrum. At this time there was a great confusion about the nature of light which was solved by de-broglie from where origin of theory of matter wave come into picture.

Debroglie Hypothesis

It supports dual nature of light (wave nature and particle nature). According to him the light consists of particles associated with definite amount of energy and momemtum. These particles were later named as photons.

The photon posses momentum and is given by

 ...(1)

P = momentum of one photon

λ = wavelength of wave.

h = Plank's constant = 6.62 × 10^-34 Js.

A photon is a packet of energy. It posses energy given by

 ...(2)

where c = speed of light

Debroglie relates particle property (momentum) with wave property (wavelength) i.e. he favours dual nature of light.

Electron Volt

It is the energy gained by an electron when it is accelerated through a potential difference of one volt.

1 eV = 1.6 × 10^-19 Joule.

Now from eq. (2)

 in Joule.

where λ is in Å

Properties of Photon

• Photon travels with speed of light.
• The rest mass of a photon is zero.
• There is no concept of photon conservation.
• All the photons of a particular frequency or wavelength posses the same energy irrespective of the intensity of the radiation.
• The increase in the intensity of the radiation imply an increase in the number of photon's crossing a given area per second.

When light travels from one medium to another medium then frequency = const (because it is the property of source) but v, λ changes

Ex.1 A beam of light having wavelength l and intensity 1 falls normally on an area A of a clean surface then find out the number of photon incident on the surface. 

Sol. Total energy incident in time t = I A t

Energy of one photon

Then number of photon incident in time t

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Electron Emission and Its Processes - Dual Nature of Radiation and Matter, Class 12, Physics

Electron Emission Process

When light is incident on a metal surface it was observed that electrons are ejected from a metal surface some times even when incredicely dim light such as that from starts and distance galaxies incident on it and some time electrons not comes out from the metal surface even high energetic or high intensity light falling on the metal surface.
This shows that the electron emission from a metal surface is not depends on the intensity of incident light but it is basically depends on the energy of the incident.
Photons no matters in number of photons are very less in a dim light, photo electric effect can be seen.
During the phenomenon of photoelectric effect one incident photon on metal surface can eject at most only one electron.
A photon is an energy packet which is fully absorbed not partially. Thus one photon can not be absorbed by more than one electron.
The minimum amount of energy of photon required to eject an electron out of a metal surface is called work function. It is denoted by φ.

The work function depends on the nature of the metal:

• The electron emission from a metal is only depends on the work function or energy of one photons.
• But how many electrons comes out from the metal is depends on intensity of the falling light on energy of the light.
• Energy of photon incident on metal will not necessarily cause emission of an electron even if its energy is more than work function. The electron after absorption may be involved in many other process like collision etc in which it can lose energy hence the ratio of no. of electrons emitted to the no. of photons incident on metal surface is less than unity.

Three Major Features of the Phtoelectric effect cannot be explained in terms of the classical of the wave theory of light.

• The intensity problem: Wave theory requires that the oscillating electric field vector E of the light wave increases in amplitude as the intensity of the light beam is increased. Since the force applied to the electron is eE, this suggests that the kinetic energy of the photoelectrons should also increased the light beam is made more intense. However observation shows that maximum kinetic energy is independent of the light intensity.
• The frequency problem: According to the wave theory, the photoelectric effect should occur for any frequency of the light, provided only that the light is intense enough to supply the energy needed to eject the photoelectrons. However observations shows that there exists for each surface a characterstic cutoff frequency n_th, for frequency less than n_th, the photoelectric effect does not occur, no matter how intense is light beam.
• The time delay problem: If the energy acquired by a photoelectron is absorbed directly from the wave incident on the metal plate, the "effective target area" for an electron in the metal is limited and probably not much more than that of a circle of diameter roughly equal to that of an atom. In the classical theory, the light energy is uniformly distributed over the wavefront. Thus, if the light is feeble enough, there should be a measurable time lag, between the impinging of the light on the surface and the ejection of the photoelectron. During this interval the electron should be absorbing energy from the beam until it had accumulated enough to escape. However, no detectable time lag has ever been measured.
  Now, quantum theory solves these problems in providing the correct interpretation of the photoelectric effect.

Threshold frequency and Threshold wavelength

We have discussed that to start photoelectric emission the energy of incident photon on metal surface must be more than the work function of the metal. If f is the work function of the metal then there must be a minimum frequency of the incident light photon which is just able to eject the electron from the metal surface. This minimum frequency or threshold frequency v_th can be given as
Threshold frequency n_th is a characteristic property of a metal as it is the minimum frequency of the light radiation required to eject a free electron from the metal surface.
As the threshold frequency is defined, we can also define threshold wavelength for a metal surface. Threshold wavelength is also called cut off wavelength. For a given metal surface threshold wavelength is the longest wavelength at which photo electric effect is possible. Thus
we have

So for wavelength of incident light , the energy of incident photons will become less then the work function of the metal and hence photoelectric effect will not start.

Thus for a given metal surface photoelectric emission will start at  or .

Einstein relation

Einstein suggested that the energy of photon  which is more than work function of a metal when incident on the metal surface is used by the electron after absorption in two parts.

(i) A part of energy of absorbed photon is used by the free electron in work done in coming out from the metal surface as work function.

(ii) The remaining part of the photon energy will be gained by the electron in the form of kinetic energy after ejection from the metal surface.

If a light beam of frequency n (each photon energy = hn) is incident on a metal surface having work function f then for , we have

 ...(1)

In equation (1) the second terms on right hand side of equation is , which is the maximum kinetic energy of the ejected electron.

In practical cases whenever an electron absorbs a photon from incident light, it comes out from the metal surface if  but in process of ejection it may collide with the neighbouring electrons and before ejection it may loose some energy during collisions with the neighbouring electrons. In this case after ejection the kinetic energy of ejected electrons will be certainly less then . If we assume there are some electrons which do not loose any energy in the process of ejection, will come out from the metal surface with the maximum kinetic energy given as

Thus all the ejected electrons from the metal surface may have different kinetic energies, distributed from 0 to .

Graph between K_max and f

Let us plot a graph between maximum kinetic energy K_max of photoelectrons and frequency f of incident light. The equation between K_max and f is,

comparing it with y = mx + c, the graph between K_max and f is a straight line with positive slope and negative intercept.

From the graph we can note the following points:

(i) K_max = 0 at f = f₀

(ii) Slope of the straight line is h, a universal constant. i.e., if graph is plotted for two different metals 1 and 2, slope of both the lines is same.

(iii) The negative intercept of the line is W, the work function, which is characteristic of a metal, i.e., intercepts for two different metals will be different. Further,

W₂ > W₁ Therefore, (f₀)₂ > (f₀)₁

Here f₀ = threshold frequency as W = hf₀

Ex.2 The photoelectric threshold of the photo electric effect of a certain metal is 2750 Å. Find

(i) The work function of emission of an electron from this metal,

(ii) Maximum kinetic energy of these electrons,

(iii) The maximum velocity of the electrons ejected from the metal by light with a wavelength 1800 Å.

Sol. (i) Given that the threshold wavelength of a metal is . Thus work function of metal can be given as

(ii) The energy of incident photon of wavelength 1800 Å on metal in eV is

Thus maximum kinetic energy of ejected electrons is

 = 6.9 - 4.52 eV = 2.38 eV

(iii) If the maximum speed of ejected electrons is v_max then we have

or  

Ex.3 Light quanta with a energy 4.9 eV eject photoelectrons from metal with work function 4.5 eV. Find the maximum impulse transmitted to the surface of the metal when each electrons flies out.

Sol. According to Einstein's photoelectric equation

If E be the energy of each ejected photo electron

momentum of electrons is

We know that change of momentum is impulse. Here the whole momentum of electron is gained when it is ejected out thus impulse on surface is

Substituting the values, we get 

Ex.4 In a experiment tungsten cathode which has a threshold 2300 Å is irradiated by ultraviolet light of wavelength 1800 Å. Calculate

(i) Maximum energy of emitted photoelectron and

(ii) Work function for tungsten

(Mention both the results in electron-volts)

Given Plank's constant joule-sec, joule and velocity of light 

Sol. The work function of tungsten cathode is

= 5.4 eV

The energy in eV of incident photons is

The maximum kinetic energy of ejected electrons can be given as

Ex.5 Light of wavelength 1800 Å ejects photoelectrons from a plate of a metal whose work functions is 2 eV. If a uniform magnetic field of tesla is applied parallel to plate, what would be the radius of the path followed by electrons ejected normally from the plate with maximum energy.

Sol. Energy of incident photons in eV is given as

As work function of metal is 2 eV, the maximum kinetic energy of ejected electrons is

 = 6.9 - eV = 4.9 eV

If v_max be the speed of fasted electrons then we have

or  

When an electron with this speed enters a uniform magnetic field normally it follows a circular path whose radius can be given by

 [As ]

or  or r = 0.149 m

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Experimental Study of Photoelectric Effect - Dual Nature of Radiation and Matter, Class 12, Physics

Experimental Study of Photo Electric Effect

Experiments with the photoelectric effect are performed in a discharge tube apparatus as illustrated in figure shown. The cathode of discharge tube is made up of a metal which shows photoelectric effect on which experiment is being carried out.

A high potential is applied to a discharge tube through a variable voltage source and a voltmeter and an ammeter are connected a measure the potential difference across the electrodes and to measure photoelectric current. Light with frequency more than threshold frequency of cathode metal is incident on it, due to which photoelectrons are emitted from the cathode. These electrons will reach the anode and constitute the photoelectric current which the ammeter will show.

Now we start the experiment by closing the switch S. Initially the variable battery source is set at zero potential. Even at zero potential variable source, ammeter will show some current because due to the initial kinetic energy some electrons will reach the anode and cause some small current will flow. But as we know majority of ejected electrons have low values of kinetic energies which are collected outside the cathode and create a could of negative charge, we call space charge, as shown in figure shown.

If the potential difference applied across the discharge tube is gradually increased from the variable source, positive potential of anode starts pulling electrons from the space charge. As potential difference increases, space charge decrease and simultaneously the photoelectric current in circuit also increases. This we can also see in the variation graph of current with potential difference as shown in figure shown.

A shown in graph, we can see as potential difference increases, current in circuit increases. But at a higher voltage V_P1 space charge vanishes and at this voltage anode is able to pull the slowest electron (zero kinetic energy) ejected by the cathode. Now as all the ejected electrons from cathode start reading anode. If further potential difference is increased, it will not make any difference in the number of electrons reaching the anode hence, further increases in potential difference will not increases the current. This we can see in figure shown that beyond V_P1 current in circuit becomes constant. This current i_s1 is called saturation current. This potential difference V_P1 at which current becomes saturated is called "pinch off voltage".

Now if the frequency of incident light is kept constant and its intensity is further increased, then the number of incident photons will increase which increases the number of ejected photo electrons so current in circuit increases and now in this case at higher intensity of incident light, current will not get saturated at potential difference V_P1 as now due to more electron emission, space charge will be more and it will not vanish at V_P1. To pull all the electrons emitted from cathode more potential difference is required. This we can se from figure shown, that at higher intensity I₂ (I₂ > I₁) current becomes saturated at higher value of potential difference V_P2.

Beyond V_P2, we can see that all the electrons ejected from cathode are reaching the anode are current become saturated at i_s2 because of more electrons. Another point we can see from figure shown that when V = 0 then also current is more at high intensity incident radiation as the number of electrons of high kinetic energy are also more in the beginning which will reach anode by penetrating the space charge.

Kinetic Energies of Electrons Reaching Anode

We know that when electrons are ejected from cathode then kinetic energies may vary from 0 to . If V is the potential difference applied across the discharge tube then it will accelerates the electron while reaching the anode. the electron which is ejected from cathode with zero kinetic energy will be the slowest one reaching the anode if its speed is v₁ at anode then we have

Similarly the electron ejected from cathode with maximum kinetic energy will be the fastest one when it will reach anode. If its speed is v₂ at anode then we have

Thus we can say that all the electrons reaching anode will have their speeds distributed from v₁ to v₂.

Reversed Potential Across Discharge Tube

Now the experiment is repeated with charging the polarity of source across the discharge tube. Now positive terminal of source is connected to the cathode of discharge tube. When a light beam incident on the cathode with , photoelectrons are ejected and move towards anode with negative polarity.

Now the electrons which are ejected with very low kinetic energy are attracted back to the cathode because of its positive polarity. Those electrons which have high kinetic energies will rush toward, anode and may constitute the current in circuit.

In this case the fastest electron ejected from cathode will be retarded during its journey to anode. As the maximum kinetic energy just after emission at cathode is , if potential difference across the discharge tube is V then the seed v_f with which electrons will reach anode can be given as

 ....(1)

Thus all the electrons which are reaching anode will have speed less then or equal to v_f. Remaining electrons which have relatively low kinetic energy will either be attracted to cathode just after ejection or will return during their journey from cathode to anode. Only those electrons will case current of flow in circuit which have high kinetic energies more then eV which can overcome the electric work against electric forces on electron due to opposite polarity of source.

Cut off Potential or Stopping Potential

We have seen with reverse polarity electrons are retarded in the discharge tube. If the potential difference is increased with reverse polarity, the number of electrons reaching anode will decrease hence photo electric current in circuit also decreases, this we can see from figure shown which shows variation of current with increase in voltage across discharge tube in opposite direction. Here we can see that at a particular reverse voltage V₀, current in circuit becomes zero. This is the voltage at which the faster electron from cathode will be retarded and stopped just before reaching the anode.

This voltage V₀, we can calculate from equation (1) by substituting v_f = 0 hence

or

or  ...(2)

or  ...(3)

We can see one more thing in figure shown that the graphs plotted for two different intensities I₁ and I₂, V₀ is same. Current in both the cases in cut off at same reverse potential V₀. The reason for this is equation-(2) and (3). It is clear that the value of V₀ depends only on the maximum kinetic energy of the ejected electrons which depends only on frequency of light and not on intensity of light. Thus in above two graphs as frequency of incident light is same, the value of V₀ is also same. This reverse potential difference V₀ at which the fastest photoelectron is stopped and current in the circuit becomes zero is called cut off potential or stopping potential.

Effect of Change in Frequency of Light on Stopping Potential

If we repeat the experiment by increasing the frequency of incident light with number of incident photons constant, the variation graph of current with voltage will be plotted as shown in figure shown.

This graph is plotted for two incident light beams of different frequency v₁ and v₂and having same photon flux. As the number of ejected photoelectrons are same in the two cases of incident light here we can see that the pinch off voltage V₀₁ as well as saturation current i_s1 are same. But as in the two cases the kinetic energy of fastest electron are different as frequencies are different, the stopping potential for the two cases will be different. In graph II as frequency of incident light is more, the maximum kinetic energy of photoelectrons will also be high and to stop it high value of stopping potential is needed. These here V₀₁ and V₀₂ can be given as

 ...(4)

and  ...(5)

In general for a given metal with work function, if V_o is the stopping potential for an incident light of frequency v then we have

or  ...(6)

or  ...(7)

Equation (7) shows that stopping potential V₀ is linearly proportional to the frequency v of incident light. The variation of stopping potential with frequency v can be shown in figure shown. Here equation .(6) can be written as

 ...(8)

This equation (8) is called Einstein's Photo Electric Effect equation which gives a direction relationship between the maximum kinetic energy stopping potential frequency of incident light and the threshold frequency.

Ex.6 Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3 V. The photo electric effect begins in this metal at frequency of . Find the work function for this metal.

Sol. The threshold frequency for the given metal surface is

Thus the work function for metal surface is

As stopping potential for the ejected electrons is 3V, the maximum kinetic energy of ejected electrons will be

According to photo electric effect equation, we have

or frequency of incident light is

Ex.7 Electrons with maximum kinetic energy 3eV are ejected from a metal surface by ultraviolet radiation of wavelength 1500 Å. Determine the work function of the metal, the threshold wavelength of metal and the stopping potential difference required to stop the emission of electrons.

Sol. Energy of incident photon in eV is

According to photo electric effect equation, we have

Þ or

or = 8.29 - 3 eV or = 5.29 eV

Threshold wavelength for the metal surface corresponding to work function 5.29 eV is given as

= 2349.9 Å

Stopping potential for the ejected electrons can be given as

Ex.8 Calculate the velocity of a photo-electron, if the work function of the target material is 1.24 eV and the wavelength of incident light is 4360 Å. What retarding potential is necessary to stop the emission of the electrons?

Sol. Energy of incident photons in eV on metal surface is

 = 2.85 eV

According to photo electric effect equation we have

 or

= 2.85 - 1.24 eV = 1.61 eV

The stopping potential for these ejected electrons can be given as

Ex.9 Determine the Planck's constant h if photoelectrons emitted from a surface of a certain metal by light of frequency 2.2 × 10¹⁵ Hz are fully retarded by a reverse potential of 6.6 V and those ejected by light of frequency 4.6 × 10¹⁵ Hz by a reverse potential of 16.5 eV.

Sol. From photo electric effect equation, we have

Here  ...(1)

and  ...(2)

Subtracting equation (1) from equation (2), we get

or

or  or = 6.6 × 10^-34 J-s

Ex.10 When a surface is irradiated with light of wavelength 4950 Å, a photo current appears which vanishes if a retarding potential greater than 0.6 volt is applied across the photo tube. When a different source of light is used, it is found that the critical retarding potential is changed to 1.1 volt. Find the work function of the emitting surface and the wavelength of second source. If the photo electrons (after emission from the surface) are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials.

Sol. In first case the energy of incident photon in eV is

 = 2.51 eV

The maximum kinetic energy of ejected electrons is

 = 0.6 eV

Thus work function of metal surface is given as

 = 2.51 - 0.6 eV = 1.91 eV

In second case the maximum kinetic energy of ejected electrons will become

 = 1.1 eV

Thus the incident energy of photons can be given as

E₂ = 1.91 + 1.1 eV = 3.01 eV

Thus the wavelength of incident photons in second case will be

 = 4129.9 Å

When magnetic field is present there will be no effect on the stopping potential as magnetic force can not change the kinetic energy of ejected electrons.

Ex.11 (a) If the wavelength of the light incident on a photoelectric cell be reduced from l₁ to l₂ Å, then what will be the change in the cut-off potential ?

(b) Light is incident on the cathode of a photocell and the stopping voltages are measured from light of two difference wavelengths. From the data given below, determine the work function of the metal of the cathode in eV and the value of the universal constant hc/e.

Wavelength (Å)                   Stopping voltage (volt)

4000                                                     1.3

4500                                                    0.9

Sol. (a) Let the work function of the surface be f. If v be the frequency of the light falling on the surface, then according to Einstein's photoelectric equation, the maximum kinetic energy KE_max of emitted electron is given by

We know that,

Where V₀= cut-off potential.

or  or

Now,

 ...(1)

(b) From equation (1), we have

 = 1.44 × 10^-6 V/m

Now,

or

or

Ex.12 A low intensity ultraviolet light of wavelength 2271 Å irradiates a photocell made of molybdenum metal. If the stopping potential is 1.3 V, find the work function of the metal. Will the photocell work if it is irradiated by a high intensity red light of wavelength 6328 Å?

Sol. The energy in eV of incident photons is

As stopping potential for ejected electrons is 1.3 V, the maximum kinetic energy of ejected electrons will be

Now from photoelectric effect equation, we have

or

or  = 4.17 eV

Energy in eV for photons for red light of wavelength 6328 Å is

As , photocell will not work if irradiated by this red light no matter however intense the light will be.

FORCE DUE TO RADIATION (PHOTON)

Each photon has a definite energy and a definite linear momentum. All photons of light of a particular wavelength l have the same energy  and the same momentum p =

When light of intensity I falls on a surface, it exerts force on that surface. Assume absorption and reflection coefficient of surface be 'a' and 'r' and assuming no transmission.

Assume light beam falls on surface of surface area 'A' perpendicularly as shown in figure.

For calculating the force exerted by beam on surface, we consider following cases.

Case (I) :

a = 1, r = 0

initial momentum of the photon =

final momentum of photon = 0

change in momentum of photon =  (upward)

energy incident per unit time = IA

no. of photons incident per unit time =Therefore, total change in momentum per unit time = n ΔP  

force on photons = total change in momentum per unit time

Therefore, force on plate due to photon(F) =  (downward)

pressure =  =  =

Case : (II)

when r = 1, a = 0

initial momentum of the photon =  (downward)

final momentum of photon =  (upward)

change in momentum

Therefore, energy incident per unit time = I A

no. of photons incident per unit time =

Therefore, total change in momentum per unit time = n. DP

force = total change in momentum per unit time

 (upward on photons and downward on the plate)

pressure

Case : (III)

When 0 < r < 1                       a + r = 1 

change in momentum of photon when it is reflected =  (upward)

change in momentum of photon when it is absorbed =  (upward)

no. of photons incident per unit time =

No. of photons reflected per unit time =

No. of photon absorbed per unit time =

force due to absorbed photon (F_a)  =  (downward)

Force due to reflected photon (F_r)  (downward)

total force = F_a + F_r

Now pressure P  =

Ex.13 Calculate force exerted by light beam if light is incident on surface at an angle q as shown in figure. Consider all cases.

Sol. Case - I a = 1, r = 0

initial momentum of photon (in downward direction at an angle q with vertical) is h/λ

final momentum of photon = 0

change in momentum (in upward direction at an angle q with vertical) =  [ ]

energy incident per unit time = I A cos θ

Intensity = power per unit normal area

 P = I A cos θ

No. of photons incident per unit time =

total change in momentum per unit time (in upward direction at an angle q with vertical)

  [ ]

Force (F) = total change in momentum per unit time

 (direction  on photon and  on the plate)

Pressure = normal force per unit Area

Pressure =  P =  =

Case II When r = 1, a = 0

Therefore, change in momentum of one photon

 (upward)

No. of photons incident per unit time

Therefore, total change in momentum per unit time

  (upward)

Therefore, force on the plate  (downward)

Pressure  

Case III 0 < r < 1, a + r = 1

change in momentum of photon when it is reflected = (downward)

change in momentum of photon when it is absorbed =  (in the opposite direction of incident beam)

energy incident per unit time = I A cos θ)

no. of photons incident per unit time =

no. of reflected photon (n_r) =

no. of absorbed photon (n_Q) =

force on plate due to absorbed photons F_a = n_a. ΔP_a

 (at an angle θ with vertical )

force on plate due to reflected photons F_r = n_r ΔP_r

 (vertically downward)

now resultant force is given by

and, pressure

Ex.14 A perfectly reflecting solid sphere of radius r is kept in the path of a parallel beam of light of large aperture. If the beam carries an intensity I, find the force exerted by the beam on the sphere.

Sol. Let O be the centre of the sphere and OZ be the line opposite to the incident beam (figure). Consider a radius about OZ to get a making an angle q with OZ. Rotate this radius about OZ to get a circle on the sphere. Change q to q +dq and rotate the radius about OZ to get another circle on the sphere. The part of the sphere between these circles is a ring of area 2pr² sinq dq. Consider a small part DA of this ring at P. Energy of light falling on this part in time Dt is

ΔU = IΔt(ΔA cos θ)

The momentum of this light falling on ΔA is ΔU/c along QP. The light is reflected by the sphere along PR. The change in momentum is

Δp = 2  cosθ =  Δt (ΔA cos² q) (direction along )

The force on ΔA due to the light falling on it, is

 ΔA cos² θ (direction along )

The resultant force on the ring as well as on the sphere is along ZO by symmetry. The component of the force on DA along ZO

cos q =  IΔA cos² θ (along )

The force acting on the ring is  dF =  I(2πr² sinθ dθ) cos³ θ

The force on the entire sphere is F =  cos³ θ dθ

F =  cos³ θ d(cos θ) = -

Note that integration is done only for the hemisphere that faces the incident beam.

De-broglie wavelength of matter wave

A photon of frequency v and wavelength λ has energy.

E = hv =

By Einstein's energy mass relation, E = mc² the equivalent mass m of the photon is given by.

m =  . ..(i)

or λ =  or λ =  ...(ii)

Here p is the momentum of photon. By analogy de-Broglie suggested that a particle of mass m moving with speed v behaves in some ways like waves of wavelength λ (called de-Broglie wavelength and the wave is called matter wave) given by,

λ =  =  ...(iii)

where p is the momentum of the particle. Momentum is related to the kinetic energy by the equation,

p =

and a charge q when accelerated by a potential difference V gains a kinetic energy K = qV. Combining all these relations Eq. (iii), can be written as,

λ=  (de=Broglie wavelength)....(iv)

de-Broglie wavelength for an electron

If an electron (charge = e) is accelerated by a potential of V volts, it acquires a kinetic energy,

K = eV

Substituting the value of h, m and q in Eq. (iv), we get a simple formula for calculating de-Broglie wavelength of an electron.

λ(in Å) =

de-Broglie wavelength of a gas molecule

let us consider a gas molecule at absolute temperature T. Kinetic energy of gas molecule is given by

K.E. =  kT ; k = Boltzman constant

λ_{gas molecules} =