The Stefan Boltzmann law describes the power radiated from a black body in terms of its temperature. Specifically, the Stefan Boltzmann law states that the total energy radiated per unit surface area of a black body across all wavelengths per unit time, is directly proportional to the fourth power of the black body's thermodynamic              temperature T
E=eσT4
where σ= =5.670373×10−8 Wm−2K−4 , Stefan Boltzmann constant.

E is the total energy radiated per unit surface area

T is the temperature of the object.

e is the emissivity (how good of a radiator/absorber the object is). The emissivity e is a correction for an approximate black body radiator, where e = 1 – R, is the fraction of the light reflected (R) by the black body. For a true black body R = 0 and e = 1 or total absorption. Which is less than or equal to 1; in the case of a blackbody e =1.

Derivation of Stefan Boltzmann Law

The total power radiated per unit area over all wavelengths of a black body can be obtained by integrating Plank’s radiation formula. Thus, the radiated power per unit area as a function of wavelength is:

  ……….(1)

• P is Power radiated.
• A is the surface area of a blackbody.
• λ is the wavelength of emitted radiation.
• h is Planck’s constant
• c is the velocity of light
• k is Boltzmann’s constant
• T is temperature.

On simplifying Stefan Boltzmann equation, we get:

On integrating both the sides with respect to λ and applying the limits we get

The integrated power after separating the constants is:

     ……….(2)

This can be solves analytically by substituting:

x =          ……….(3)

Therefore, dx= d

d = d        ……….(4)

Moreover h and c can also be written in terms of x as

h =          ……….(5)

c =          ……….(6)

Substituting the above equation (3),(4),(5) and (6) in equation (2)

= 2π (()2

= 2π (

= 2π (        ……….(7)

The above equation can be comparable to the standard form of integral:

=          ……….(8)

So from equation (8) we can rewrite equation (7) as

= 2π (

=  (        ……….(9)

On further simplifying we get,

 =  σ         ……….(10)

Thus, we arrive at a mathematical form of Stephen Boltzmann law:

⇒ E = σ         ……….(11)

Where,

E =

Where σ= =5.670373×10−8 Wm−2K−4 , Stefan Boltzmann constant.

So this is the mathematical derivation of Stefan’s Law.

Example: A body of emissivity (e = 0.75), the surface area of 300 cm2 and temperature 227 ºC are kept in a room at temperature 27 ºC. Using the Stephens Boltzmann law, calculate the initial value of net power emitted by the body.

Solution:

E = = eσ(T4 – T04)

P = eσA(T4 – T04)

P  = (0.75) (5.67 × 10-8 W/m2 – k4) (300 × 10-4 m2) × [(500 K)4 – (300 K)4]

= 69.4 Watts.

Wien’s law also called Wien’s displacement law. It is named after German physicist Wilhelm Wien, who received the Nobel Prize for Physics in 1911 for discovering the law.

This law gives relationship between the temperature of a blackbody and the wavelength at which it emits the most light.

Wien found that the radiative energy dW per wavelength interval dλ has a maximum at a certain wavelength λm and that the maximum shifts to shorter wavelengths as the temperature T is increased. He found that the product λmT is an absolute constant.

Mathematical representation of the law:

λmax =bT

Where, b is the Wien’s displacement constant =2897.8 µm K

T is the temperature in kelvins

Wien’s constant: b (Wien’s displacement constant)

Physical constant defining the relationship between the thermodynamic temperature of the black body and the wavelength is known as Wien’s constant. It is a product of temperature and wavelength of the black body which grows shorter as the wavelength reaches a maximum with temperature.

The spectral distribution as a function of temperature is now to be examined more closely. It turns out that the maximum of the curve shifts with increasing temperature to ever shorter wavelengths. The dependence of this wavelength λmax. on the temperature is given by the following equation. This equation is also known as Wien’s displacement law.

λmax. = 2897.8 µm K / T   Wien’s displacement law

The Wien’s displacement law can be obtained by determining the maxima of Planck’s law. For this purpose, the function must be derived with respects to the wavelength λ.

By using the product rule and setting the derivative equal to zero, we get

This equation will only be zero if the term in the round bracket becomes zero

This equation can only be solved numerically, e.g. with the Newton’s method. The result will be x = 4,9651. With this result the wavelength λmax can be determined as a function of the temperature by rearranging the equation.

The maximum of the spectral intensity can also be determined for the frequency form Is(f). For this the function Is(f) must be derived with respect to frequency f and setting the derivative equal to zero:

Example: A hot black body emits the energy at the rate of 16 J m-2 s-1 and its most intense radiation corresponds to 20,000 Å. When the temperature of this body is further increased and its most intense radiation corresponds to 10,000 Å, then find the value of energy radiated in Jm-2 s-1.

Solution:

Wein’s displacement law is, λm.T = b

i.e. T 1/ λm

so T1/T2 = λ2/λ1

so T1/T2 = 10,000/20,000 = 1/2

Here, λm   becomes half, the Temperature doubles.

Now from Stefan Boltzmann Law, e = sT4

e1/e2 = (T1/T2)4

⇒ e2 = (T2/T1)4 . e1 = (2)4 . 16

= 16.16 = 256 J m-2 s-1

Example: Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures T1 and T2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B ?

Solution:

Stefan-Boltzmann law gives the energy radiated per unit time by a spherical black body of area A=4πr2 and temperature T

E=σAT4=4πσr2T4.

The Wien's displacement law relates temperature of the black body to the wavelength at maximum intensity by λmT=b.

Eliminate T from above equations to get, E=4πσb4(r2/λ4), which gives,

=( ( =( ( =   9.

RAYLEIGH-JEANS LAW

In his 1900 attempt, Rayleigh focused on understanding the nature of the electromagnetic radiation inside the cavity. He considered the radiation to consist of standing waves having a temperature T with nodes at the metallic surfaces. These standing waves, he argued, are equivalent to harmonic oscillators, for they result from the harmonic oscillations of a large number of electrical charges, electrons that are present in the walls of the cavity. When the cavity is in thermal equilibrium, the electromagnetic energy density inside the cavity is equal to the energy density of the charged particles in the walls of the cavity; the average total energy of the radiation leaving the cavity can be obtained by multiplying the average energy of the oscillators by the number of modes (standing waves) of the radiation in the frequency interval ν to ν +d ν.

The Rayleigh-Jeans Radiation Law was a useful, but this law is not completely successful attempt at establishing the functional form of the spectra of thermal radiation.

The energy density uν per unit frequency interval at a frequency ν is, according to the The Rayleigh-Jeans Radiation,

uν =8πν2kTc2         …………….(1)

Where k is Boltzmann's constant, T is the absolute temperature of the radiating body and c is the speed of light in a vacuum.

This formula fits the empirical measurements for low frequencies, but fails increasingly for higher frequencies. The failure of the formula to match the new data was called the ultraviolet catastrophe.

Derivation

Consider a cube of edge length L in which radiation is being reflected and re-reflected off its walls. Standing waves occur for radiation of a wavelength λ only if an integral number of half-wave cycles fit into an interval in the cube. For radiation parallel to an edge of the cube this requires

=m  ……….(1)

Where m is an integer

=       ……….(2)

Between two end points there can be two standing waves, one for each polarization. In the following the matter of polarization will be ignored until the end of the analysis and there the number of waves will be doubled to take into account the matter of polarization.

Since the frequency ν is equal to c/λ, where c is the speed of light

  ……….(3)

It is convenient to work with the quantity q, known as the wavenumber, which is defined as

……….(4)

and hence

  ……….(5)

In terms of the relationship for the cube,

q = = π ( )   ……….(6)

and hence

q2 = π2  ( )2      ……….(7)

Another convenient term is the radian frequency ω=2πν. From this it follows that q=ω/c.

If mX, mY, mZ denote the integers for the three different directions in the cube then the condition for a standing wave in the cube is that

q2 = π2 [()2 + ()2 + ()2]   ……….(8)

This reduces to

m2x + m2y + m2z =    ……….(9)

Now the problem is to find the number of non-negative combinations of (mX, mY, mZ) that fit between a sphere of radius R and and one of radius R+dR. First the number of combinations ignoring the non-negativity requirement can be determined.

The volume of a spherical shell of inner radius R and outer radius R+dR is given by

dV=4πR2dR   ……….(10)

if

R =      ……….(11)

then

R = =       ……….(12)

and hence

dR =        ……….(13)

this means that

dV = 4π(2Lν/c)2(2L/c) dν = 32π(L3v2/c3)dν  ……….(14)

Now the non-negativity require for the combinations (mX, mY, mZ) must be taken into account.

We know that the nonnegative combinations are in one quadrant of circle for the two dimensional case. The approximation arises from the matter of the combinations on the boundaries of the nonnegative quadrant. For the three dimensional case the nonnegative combinations constitute approximately one octant of the total. Thus the number dN for the nonnegative combinations of (mX, mY, mZ) in this volume is equal to (1/8)dV and hence

dN = 4πv2dν     ……….(15)

The average kinetic energy per degree of freedom is ½kT, where k is Boltzmann's constant. For harmonic oscillators there is an equality between kinetic and potential energy so the average energy per degree of freedom is kT. This means that the average radiation energy E per unit frequency is given by

= kT () = 4πkT( )v2    ……….(16)

and the average energy density, uν, is given by

=  () () =      ……….(17)

The previous only considered one direction of polarization for the radiation. If the two directions of polarization are taken into account a factor of 2 must be included in the above formula; i.e.,

=       ……….(18)

This is the Raleigh-Jeans Law of Radiation and holds empirically as the frequency goes to zero.

The Rayleigh-Jeans Law was an important step in our understanding of the equilibrium radiation from a hot object, even though it turned out not to be an accurate description of nature. The careful work in developing the Rayleigh-Jeans law laid the foundation for the quantum understanding expressed in the Planck radiation formula. By Rayleigh-Jeans Law the energy density of radiation diverges at high frequencies.

This is the famous result known as the ultraviolet catastrophe– the total energy in black-body radiation diverges. This expression for the spectral energy distribution of radiation is known as the Rayleigh-Jeans Law and, although it diverges at high frequencies, it is in excellent agreement with the measured spectrum at low frequencies and high temperatures.

Derivation of Planck’s law

While Wien’s formula and the Rayleigh-Jeans Law could not explain the spectrum of a black body, Max Planck’s equation solved the problem by assuming that light was discrete.

The photoelectric effect demonstrates that light waves have particle properties and that the light quanta or photons of a particular frequency v each have energy hv. We need to reconcile this picture with the classical picture of electromagnetic waves in a box. In the classical picture, the energy associated with the waves is stored in the oscillating electric and magnetic fields.

We found it necessary to impose the constraint that only certain modes are permitted by the boundary conditions; the waves are constrained to fit into the box with whole numbers of half wavelengths in the x, y, z directions.

Now we have a further constraint. The quantisation of electromagnetic radiation means that the energy of a particular mode of frequency v  cannot have any arbitrary value but only those energies which are multiples of hν, in other words the energy of the mode is E(ν) =nhv, where we associate n photons with this mode.

We now consider all the modes (and photons) to be in thermal equilibrium at temperature T. In order to establish equilibrium, there must be ways of exchanging energy between the modes (and photons) and this can occur through interactions with any particles or oscillators within the volume or with the walls of the enclosure. We now use the Boltzmann distribution to deter-mine the expected occupancy of the modes in thermal equilibrium. The probability that a single mode has energy En=nhv is given by the usual Boltzmann factor

         ……….(19)

Where the denominator ensures that the total probability is unity, the usual normalisation procedure. In the language of photons, this is the probability that the state contains n photons of frequency ν. The mean energy of the mode of frequency ν is therefore

……….(20)

To simplify the calculation, let us substitute x=exp(−hν/kT). Then (20) becomes

……….(21)

Now, we remember the following series expansions:

   ………….(22)

Hence, the mean energy of the mode is

   ………….(23)

This is the result we have been seeking. To find the classical limit, we allow the energy quanta hν to tend to zero.

Expanding ehν/kT−1 for small values of hν/kT,

ehν/kT−1 = 1 +hν/kT+!(hν/kT)2+···−1      ………….(24)

Thus, for small values of hν/kT, ehν/kT−1 =hν/kT.

And also

     …………….(25)

Thus, if we take the classical limit, we recover exactly the expression for the average energy of a harmonic oscillator in thermal equilibrium, =kT. We can now complete the determination of Planck’s radiation formula.

We have already shown that the number of modes in the frequency interval ν to ν+dν is (8πν2/c3) dν per unit volume. The Planck distribution in terms of the energy density of radiation per unit frequency interval

The energy density of radiation in this frequency range is

   ………….(26)

This is the Planck distribution function.

As we know in the Photoelectric Effect, the Compton Effect, and the pair production effect—radiation exhibits particle-like characteristics in addition to its wave nature. In 1923 de Broglie took things even further by suggesting that this wave–particle duality is not restricted to radiation, but must be universal.

In 1923, the French physicist Louis Victor de Broglie (1892-1987) put forward the bold hypothesis that moving particles of matter should display wave-like properties under suitable conditions.

All material particles should also display dual wave–particle behaviour. That is, the wave–particle duality present in light must also occur in matter.

So, starting from the momentum of a photon p = hν/c = h/λ.

We can generalize this relation to any material particle with nonzero rest mass.  Each material particle of momentum  behaves as a group of waves (matter waves) whose wavelength λ and wave vector  are governed by the speed and mass of the particle. De Broglie proposed that the wave length λ associated with a particle of momentum p is given as where m is the mass of the particle and v its speed.

λ =   =    …….(1)    =     …….(2) 

Where ℏ = h/2π. The expression known as the de Broglie relation connects the momentum of a particle with the wavelength and wave vector of the wave corresponding to this particle. The wavelength λ of the matter wave is called de Broglie wavelength. The dual aspect of matter is evident in the de Broglie relation.

λ is the attribute of a wave while on the right hand side the momentum p is a typical attribute of a particle. Planck’s constant h relates the two attributes. Equation (1) for a material particle is basically a hypothesis whose validity can be tested only by experiment.

However, it is interesting to see that it is satisfied also by a photon. For a photon, as we have seen, p = hν/c.

Therefore

= = λ

Example: What is the frequency of a photon with energy of 4.5 eV?

Solution:

E = (4.5 eV) x (1.60 x 10-19 J/eV) = 7.2 x 10-19 J

E = hf

h = 6.63 x 10-34 J x s

f = E / h = (7.2 x 10-19 J) / (6.63 x 10-34 J x s)

f = 1.1   x 1015 Hz

De Broglie’s Hypothesis: Matter Waves

Matter waves: According to De-Broglie, a wave is associated with each moving particle which is called matter waves.

As we know in the photoelectric effect, the Compton Effect, and the pair production effect radiation exhibits particle-like characteristics in addition to its wave nature. In 1923 de Broglie took things even further by suggesting that this wave–particle duality is not restricted to radiation, but must be universal:

all material particles should also display a dual wave–particle behaviour.

That is, the wave–particle duality present in light must also occur in matter. So, starting from the momentum of a photon p = h ν /c = h/λ, we can generalize this relation to any material particle with nonzero rest mass: each material particle of momentum behaves as a group of waves (matter waves) whose wavelength λ and wave vector are governed by the speed and mass of the particle.

λ =       =

Wave has wavelength λ here h is Planck's constant and p is the momentum of the moving particle.
 

We have seen that microscopic particles, such as electrons, display wave behaviour. What about macroscopic objects? Do they also display wave features? They surely do. Although macro-scopic material particles display wave properties, the corresponding wavelengths are too small to detect; being very massive, macroscopic objects have extremely small wavelengths.

At the microscopic level, however, the waves associated with material particles are of the same size or exceed the size of the system. Microscopic particles therefore exhibit clearly noticeable wave-like aspects.

The general rule is: whenever the de Broglie wavelength of an object is in the range of, or exceeds, its size, the wave nature of the object is detectable and hence cannot be neglected.

But if its de Broglie wavelength is much too small compared to its size, the wave behaviour of this object is undetectable.

For a quantitative illustration of this general rule, let us calculate in the following example the wavelengths corresponding to two particles, one microscopic (electron) and the other macroscopic(ball).

Example: What is the de Broglie wavelength associated with (a) an electron moving with a speed of 5.4×106 m/s, and (b) a ball of mass 150 g travelling at 30.0 m/s?

Solution:

(a)For the electron:

 Mass m = 9.11×10–31 kg, speed v = 5.4×106 m/s.

Then, momentum

p = m v = 9.11×10–31 kg × 5.4 × 106 (m/s)

p = 4.92 × 10–24 kg m/s

de Broglie wavelength, λ = h/p = 6.63 x 10-34Js/ 4.92 × 10–24 kg m/s

λ= 0.135 nm

(b)For the ball:

Mass m’ = 0.150 kg,

Speed v ’= 30.0 m/s.

Then momentum p’ = m’ v’= 0.150 (kg) × 30.0 (m/s)

p’= 4.50 kg m/s

de Broglie wavelength λ’ = h/p’ =6.63 x 10-34Js/ 4.50 kg m/s =1.47 ×10–34 m

The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10–19 times the size of the proton, quite beyond experimental measurement.

Schrodinger wave equation, is the fundamental equation of quantum mechanics, same as the second law of motion is the fundamental equation of classical mechanics. This equation has been derived by Schrodinger in 1925 using the concept of wave function on the basis of de-Broglie wave and plank’s quantum theory.

Let us consider a particle of mass m and classically the energy of a particle is the sum of the kinetic and potential energies. We will assume that the potential is a function of only x.

 So We have

E= K+V =mv2+V(x) = +V(x)   ……….. (1)

By de Broglie’s relation we know that all particles can be represented as waves with frequency ω and wave number k, and that E= ℏω and p= ℏk.

Using this equation (1) for the energy will become

ℏω = + V(x)    ……….. (2)

A wave with frequency ω and wave number k can be written as usual as

ψ(x, t) =Aei(kx−ωt)     ……….. (3)

the above equation is for one dimensional and for three dimensional we can write it as

ψ(r, t) =Aei(k·r−ωt)    ……….. (4)

But here we will stick to one dimension only.

=−iωψ   ⇒   ωψ=   ……….. (5)   

=−k2ψ  ⇒   k2ψ = -    ……….. (6)

If we multiply the energy equation in Eq. (2) by ψ, and using(5) and (6) , we obtain

ℏ(ωψ)  = ψ+ V(x) ψ  ⇒   = - + V(x) ψ  ……….. (7)

This is the time-dependent Schrodinger equation.

If we put the x and t in above equation then equation (7)  takes the form as given below

= - + V(x) ψ(x,t)    ……….. (8)

In 3-D, the x dependence turns into dependence on all three coordinates (x, y, z) and the  term becomes ∇2ψ.

The term |ψ(x)|2 gives the probability of finding the particle at position x.

Let us again take it as simply a mathematical equation, then it’s just another wave equation. However We already know the solution as we used this function ψ(x, t) =Aei(kx−ωt) to produce Equations (5), (6) and (7)

But let’s pretend that we don’t know this, and let’s solve the Schrodinger equation as if we were given to us. As always, we will guess an exponential solution by looking at exponential behaviour in the time coordinate, our guess is ψ(x, t) =e−iωtf(x) putting this into Equation (7) and cancelling the e−iωt yields

  = - + V(x) f(x)   ……….. (9)

We already know that E=. However ψ(x, t) is general convention to also use the letter ψ to denote the spatial part. So we will now replace f(x) with ψ(x)

   Eψ = - + V(x) ψ    ……….. (10)

This is called the time-independent Schrodinger equation.

It is not possible to measure all properties of a quantum system precisely. Max Born suggested that the wave function was related to the probability that an observable has a specific value.

 In any physical wave if ‘A’ is the amplitude of the wave, then the energy density i.e., energy per unit volume is equal to ‘A2’. Similar interpretation can be made in case of mater wave also. In matter wave, if ‘Ψ ‘is the wave function of matter waves at any point in space, then the particle density at that point may be taken as proportional to ‘Ψ2’ . Thus Ψ2 is a measure of particle density.

According to Max Born ΨΨ*=Ψ2 gives the probability of finding the particle in the state ‘Ψ’. i.e., ‘Ψ2’ is a measure of probability density. The probability of finding the particle in a volume( dv=dxdydz) is given by

=

 Since the particle has to be present somewhere, total probability of finding the particle somewhere is unity i.e., particle is certainly to be found somewhere in space. i.e

=1

Or  =1

This condition is called Normalization condition. A wave function which satisfies this condition is known as normalized wave function.

The wave function, at a particular time, contains all the information that anybody at that time can have about the particle.  But the wave function itself has no physical interpretation.  It is not measurable.  However, the square of the absolute value of the wave function has a physical interpretation.  We interpret |ψ(x,t)|2 as a probability density, a probability per unit length of finding the particle at a time t at position x.

The wave function ψ associated with a moving particle is not an observable quantity and does not have any direct physical meaning. It is a complex quantity. The complex wave function can be represented as

 ψ(x, y, z, t) = a + ib

and its complex conjugate as

ψ*(x, y, z, t) = a – ib.

The product of wave function and its complex conjugate is

ψ(x, y, z, t)ψ*(x, y, z, t) = (a + ib) (a – ib) = a2 + b2

a2 + b2 is a real quantity.

However, this can represent the probability density of locating the particle at a place in a given instant of time.

The positive square root of ψ(x, y, z, t) ψ*(x, y, z, t) is represented as |ψ(x, y, z, t)|, called the modulus of ψ. The quantity |ψ(x, y, z, t)|2 is called the probability. This interpretation is possible because the product of a complex number with its complex conjugate is a real, non-negative number. 

We should be able to find the particle somewhere, we should only find it at one place at a particular instant, and the total probability of finding it anywhere should be one. 

For the probability interpretation to make sense, the wave function must satisfy certain conditions. 

• The wave function must be single valued at each point.
• The probability of finding the particle at time t in an interval ∆x must be some number between 0 and 1.
• ψ must be finite everywhere.
• ψ must be continuous everywhere and must also be continuous everywhere except where V(x) is infinite.
• ψ (x) must vanish ψ0 as x.
• The wave function should satisfy the normalization condition. Normalization condition of a wave function ψ is mathematical statement of existence of the particle somewhere.  so that if we sum up all possible values ∑|ψ(xi,t)|2∆xi  we must obtain 1.  The total probability of finding the particle anywhere must be one. Normalization condition is given as

dx =1

Only wave function with all these properties can yield physically meaningful   result.

Physical significance of wave function

• The wave function ‘Ѱ’ has no physical meaning. it is a complex quantity representing the variation of a matter wave.
• The wave function Ѱ(r,t) describes the position of particle with respect to time .
• It can be considered as ‘probability amplitude’ since it is used to find the location of the particle.
• The square of the wave function gives the probability density of the particle which is represented by the wave function itself.
• More the value of probability density, more likely to find the particle in that region.

The Schrodinger equation also known as Schrodinger’s wave equation is a partial differential equation that describes the dynamics of quantum mechanical systems by the wave function. The trajectory, the positioning, and the energy of these systems can be retrieved by solving the Schrodinger equation.

All of the information for a subatomic particle is encoded within a wave function. The wave function will satisfy and can be solved by using the Schrodinger equation. The Schrodinger equation is one of the fundamental axioms that are introduced in undergraduate physics.

Particle in a One Dimensional Deep Potential Well

Let us consider a particle of mass ‘m’ in a deep well restricted to move in a one dimension (say x). Let us assume that the particle is free inside the well except during collision with walls from which it rebounds elastically.

The potential function is expressed as

V= 0 for 0      ………. (1)

V= for x <0, x>L     ………. (2)

Figure : Particle in deep potential well

The probability of finding the particle outside the well is zero (i.e. Ѱ =0) 

 Inside the well, the Schrödinger wave equation is written as

ψ + E ψ =0    …………….(2)    

Substituting   E = k2                                       …………….(3)

writing the SWE for 1-D we get

+ k2 ψ =0       …………….(4)

The general equation of above equation may be expressed as

ψ = Asin (kx + ϕ)      …………….(5)

Where A and ϕ  are constants to be determined by boundary conditions

Condition I: We have ψ = 0 at x = 0, therefore from equation

0 = A sinϕ   

As A then sinϕ =0  or  ϕ=0    …………….(6)

Condition II: Further ψ = 0 at x = L, and ϕ=0 , therefore  from equation (5)

0 = Asin kL

As A then sinkL =0  or  kL=nπ

k =       …………….(7)

where n= 1,2,3,4………

Substituting the value of k from (7) to (3)

)2 = E

 This gives energy of level  

En =    n=1,2,3,4…so on    …………….(8)

 
From equation En is the energy value (Eigen Value) of the particle in a well.

It is clear that the energy values of the particle in well are discrete not continuous.

Using (6) and (7) equation (5) becomes, the corresponding wave functions will be

ψ = ψn = Asin   …………….(9)

The probability density

|ψ(x,t)|2 = ψ ψ*

|ψ(x,t)|2 = A2sin2   …………….(10)

The probability density is zero at x = 0 and x = L. since the particle is always within the well

    …………….(11)

=1

A =

Substituting A in equation (9) we get

ψ = ψn = sin    n=1,2,3,4…..    …………….(12)

The above equation (12) is normalized wave function  or Eigen function belonging to energy value En

Figure: Wave function for Particle

A Free Particle

A particle is said to be free when no external force is acting on during its motion in the given region of space, and its potential energy V is constant.

Let us consider an electro is freely moving in space in positive x direction and not acted by any force, there potential will be zero. The Schrodinger wave equation reduces to

ψ + E ψ =0        

Substituting   E = k2                             

As the electron is moving in one direction (say x axis), then the above equation can be written as

+ k2 ψ =0 

The general solution of the equation (2) is of the form ψ = ψ0e-iωt  

The electron is not bounded and hence there are no restrictions on k. This implies that all the values of energy are allowed. The allowed energy values form a continuum and are given by

E =

The wave vector k describes the wave properties of the electron.

 It is seen from the relation that E k2 Thus the plot of E as a function of k gives a parabola.

The momentum is well defined and in this case given by

px ψ =

Therefore, according to uncertainty principle it is difficult to assign a position to the electron.

Here we will discuss elastic scattering of a photon from a free electron as shown in figure. Consider that the incident photon, of energy E =hν and momentum p = hν /c, collides with an electron that is initially at rest. If the photon scatters with a momentum at an angle θ while the electron recoils with a momentum the conservation of linear momentum yields

= +                    …………(1)

Which leads

= ( - )2 = + -2pp’cosθ = ( + -2νν’cosθ)   ………..(2)

Energy Conservation

The energies of the electron before and after the collision are given, respectively, by

E0 =mec2       ………..(3)

Since the energies of the incident and scattered photons are given by E = hν and E0 = hν’, respectively, conservation of energy dictates that

E + E0  = E’ + Ee        ………..(4)             ………..(5)

Which leads to

  ………..(6)

Squaring both sides of (5) and simplifying, we get

Hence wavelength shift is given by

Where λC = h/mec = 2.426 x 10-12 m is called the Compton wavelength of the electron.

This relation connects the initial and final wavelengths to the scattering angle.

Example: High energy photons are scattered from electrons initially at rest. Assume the photons are back scattered and their energies are much larger than the electron’s rest-mass energy, E = mec2.

(a) Calculate the wavelength shift.

(b) Show that the energy of the scattered photons is half the rest mass energy of the electron, regardless of the energy of the incident photons.

(c) Calculate the electron’s recoil kinetic energy if the energy of the incident photons is 150 MeV.

Solution:

(a)  In the case where the photons backscatter i.e. θ = π

The wavelength shift becomes

= - = 2c sin2θ = 2c sin2 π = 2c = 2 x 2.426 x 10-12 m =4.86 x 10-12 m

(b) Since the energy of the scattered photons E’ is related to the wavelength   by E =hc / 

Where E = hc/  is the energy of the incident photons. If E = mec2 we can approximate

(c)  If E = 150 MeV, the kinetic energy of the recoiling electrons can be obtained from conservation of energy

Ke = E – E’ 150 – 0.25 =14.75 MeV

Example: X-rays with an  energy of 300 keV undergo Compton scattering with a target. If the scattered X-rays are detected at 30° relative to the incident X-rays, determine the Compton shift at this angle, the energy of the scattered X-ray, and the energy of the recoiling electron.

Solution:

 The Compton shift

When it is scattered through an angle θ by an electron: λ′−λ= λe(1−cosθ)

We know Compton wavelength of the electron λe= h/mec= 2.43 pm

me mass of the electron = 511 keV/c2

θ = 30°

Compton shift is

λ′−λ= λe(1−cosθ) = 2.426 x 10-12 m (1−cos30◦.) = 0.325 pm

The energy E′ of the scattered photon is E′=hc/ λ′

And λ=hc/E

E=300 keV is the wavelength of the incoming photon. It follows that E′= 278 keV. By conservation of energy, the energy lost by the photon in the collision is converted into kinetic energy K of the recoiling electrons K= 22 keV