A photon is basically a basic particle which carries with itself electromagnetic energy. The light coming from sun has different wavelengths or energy and on the basis of it, we have different regions like visible, infrared, ultraviolet and many more. But one thing is common in all these regions is photon but of different frequency so different energy.

Thus photon is basically a quantum of light.

Properties:

(i) In the interaction of radiation with matter, radiation behaves as if it is made of particles like photons

(ii) Each photon has energy E= hu and momentum p = mv or p =

(iii)Irrespective of the intensity of radiation, all the photons of particular frequency or wavelength have the same energy and same momentum.

(iv)All the photons emitted from a source travels in space with the speed of light.

(v) The velocity of photon in different media is different.

(vi) The rest mass of photon is zero.

(vii) They are not deflected by electric or magnetic fields.

(viii) In a photon particle collision, the energy and momentum are conserved but the number of photons may not be conserved.

This is about photons and its properties.

Brain Teaser:

If photon has mass zero, then why it has momentum?

If you know the answer, please share with us.

The post Photon and its properties first appeared on Winner Science.

I.e Potential energy V(x) is of the form

V(x) = {o; if o<x<L

∞: elsewhere

The one-dimensional time independent Schrodinger wave equation is given by

d²ψ/dx²+ 2m/Ћ²[E-V] ψ=0                                             (1)

Here we have changed partial derivatives in to exact because equation now contains only one variable i.e x-Co-ordinate. Inside the box V(x) =0

Therefore   the Schrodinger equation in this region becomes

d²/ψ/dx²+ 2m/Ћ²Eψ=0

Or                 d²ψ/dx²+ K²ψ=0                                          (2)

Where                       k=    2mE/Ћ² (3)

K is called the Propagation constant of the wave associated with particle and it has dimensions reciprocal of length.

The general solution of eq (2) is

Ψ=A sin Kx + B cos K x                                   (4)

Where A and B are arbitrary conditions and these will be determined by the boundary conditions.

(ii) Boundary Conditions

The particle will always remain inside the box because of infinite potential barrier at the walls. So the probability of finding the particle outside the box is zero i.e.ψx=0 outside the box.

We know that the wave function must be continuous at the boundaries of potential well at x=0 and x=L, i.e.

Ψ(x)=0 at x=0                                            (5)

Ψ(x)=0 at x= L                                           (6)

These equations are known as Boundary conditions.

(iii) Determination of Energy of Particle

Apply Boundary condition of eq.(5) to eq.(4)

0=A sin (X*0) +B cos (K*0)

0= 0+B*1

B=0                                                             (7)

Therefore eq.(4) becomes

Ψ(x) = A sin Kx                                    (8)

Applying the boundary condition of eq.(6) to eq.(8) ,we have

0=A sin KL

Sin KL=0

KL=nπ

K=nπ/L                                                                       (9)

Where    n= 1, 2, 3 – – –

A Cannot be zero in eq. (9) because then both A and B would be zero. This will give a zero wave function every where which means particle is not inside the box.

Wave functions. Substitute the value of K from eq. (9) in eq. (8) to get

Ψ(x)=A sin(nπ/Lx)

As the wave function depends on quantum number π so we write it ψ_n. Thus

Ψ_n=A sin (nπx/L)0<x<L

This is the wave function or eigen function of the particle in a box.

Ψ_n=0    outside the box

Energy value or Eigen value of particle in a box: Put this value of K from equation (9) in eq. (3)

nπ/L = 2m E/Ћ²

Squaring both sides

n²π²/L²=2mE/Ћ²

E=n²π²Ћ²/2mL²

Where n= 1, 2, 3… Is called the Quantum number

As E depends on n, we shall denote the energy of particle ar E_n. Thus

E_n= n²π²Ћ²/2mL² (10)

This is the eigen value or energy value of the particle in a box.

The post Application of Schrodinger wave equation: Particle in a box first appeared on Winner Science.

If there is a wave associated with a particle, then there must be a function to represent it. This function is called wave function.

Wave function is defined as that quantity whose variations make up matter waves. It is represented by Greek symbol ψ(psi), ψ consists of real and imaginary parts.

Ψ=A+iB

PHYSICAL SIGNIFICANCE OF WAVE FUNCTIONS (BORN’S INTERPRETATION):

Born’s interpretation

The wave function ψ itself has no physical significance but the square of its absolute magnitude |ψ²| has significance when evaluated at a particular point and at a particular time |ψ²| gives the probability of finding the particle there at that time.

The wave function ψ(x,t) is a quantity such that the product

P(x,t)=ψ^*(x,t)ψ(x,t)

Is the probability per unit length of finding the particle at the position x at time t.

P(x,t) is the probability density and ψ^*(x,t) is complex conjugate of ψ(x,t)

Hence the probability of finding the particle is large wherever ψ is large and vice-versa.

NORMALIZATION CONDITION

The probability per unit length of finding the particle at position x at time t is

P=ψ^*(x,t)ψ(x,t)

So, probability of finding the particle in the length dx is

Pdx=ψ^*(x,t)ψ(x,t)dx

Total probability of finding the particle somewhere along x-axis is

∫pdx =∫ ψ^*(x,t)ψ(x,t)dx

If the particle exists , it must be somewhere on the x-axis . so the total probability of finding the particle must be unity i.e.

∫ψ^*(x,t)ψ(x,t)dx=1                               (1)

This is called the normalization condition . So a wave function ψ(x,t) is said to be normalized if it satisfies the condition(1)

ORTHOGONAL WAVE FUNCTIONS

Consider two different wave functions ψ_m and ψ_n such that both satisfies Schrodinger equation.These two wave functions are said to be orthogonal if they satisfy the conditions.

Or                        ∫ ψ_n^* (x,t) ψ_m(x,t) dV=0 for n≠m]                          ( 1)

∫ ψ_n^* (x,t) ψ_m(x,t) dV=0 for m≠n ]

If both the wave functions are simultaneously normal then

∫ ψ_m ψ_m^* d V=1=∫ψ_nψ_n^* dV                                   (2)

Orthonormal wave functions:

The sets of wave functions, which are both normalized as well as orthogonal are called orthonormal wave functions.

Equations (16) and (17) are collectively written as

∫ψ^*_mψ_ndV={ o if   m≠n

=[1 if m=n

The post Wave function and its physical significance first appeared on Winner Science.

V=A exp[-i/Ћ(Et-px]

= A exp(-i/Ћ Et) exp(i/Ћ)

Ψ=ψ^’ exp(-iEt/Ћ)                                                             (1)

Where Ћ = h/2π

So, ψ is a product of a time dependent function exp(-i/Ћ Et) and a position dependent function

Ψ^’= A exp(-i/Ћ px)

Differentiating equation (1) w.r.t.x, We have

dψ/dx = exp(-i/ЋEt) dψ^’/dx

and            d²ψ/dx²= exp(-i/Ћ Et) d²ψ^’/dx² (2)

Also on differentiating ψ w.r.t. t, we have

dψ/dt=ψ^’ exp (-iEt/Ћ) (I E/Ћ)

dψ/dt=-(iE/Ћ)ψ^’ exp(-I Et/Ћ)                                           (3)

Put equations [1-3] in time dependent Schrodinger wave equation (discussed earlier),

iЋ[-iE/Ћψ^’ exp(-iEt/Ћ)]= -Ћ²/2m[exp(i/ЋEt) d²ψ^’/dx²] +V ψ^’ exp(iEt/Ћ)

Eψ^’ exp(iEt/Ћ) = -Ћ²/2m exp(i/Ћ Et) d²ψ^’/dx² + V ψ^’exp(iEt/Ћ)

Dividing throughout by expression (i/Ћ Et) we have

Eψ^’= (-Ћ²/2π) d²ψ^’/dx²+V ψ^’

Or            (E-V)ψ^’=-Ћ²/2m  dψ^’/dx²

or             d²Ψ^’/dx² + (2m/Ћ²)(E-V)ψ^’ (4)

Which is time independent form of Schrodinger wave equation in one dimension.

In three-dimensional form:

d² Ψ^’/dx²+ d²ψ^’/dy²+ d²Ψ’/ d²x²+2m/Ћ²(E-V)ψ^’=0

In this equation, ψ’ equation, ψ’(x) is also called the wave function. The potential V(x) does not contain the time explicity and E, the total energy of the particle is a constant.

The post Time Independent Schrodinger Wave Equation first appeared on Winner Science.

Ψ(x,t)=A exp[-i(ωt-kx)]

Put ω=2πv and K=2π/h

Ψ(x,t)=A exp[ -i(2πvt-2π/h x)]                                      (1)

If E= total energy of the particle

P= momentum of the particle then

E=hv=2πЋ/p

Where Ћ = h/2π

Putting in equation (1), we get

Ψ(x,t)= A exp[-i(E/Ћ t –p/ Ћ x)]

Ψ(x,t) = A exp[-i/ Ћ(Et-px)]                     (2)

This equation is a description of the wave equivalent of a free particle moving in the +ve x- direction. But generally we are interested in situations where particle is not free i.e it is subjected to some external force.

Differentiating equation (2) w.r.t. x

dΨ/d x =A exp[-i/ Ћ(Et-px]  / x[-i/h(Et-px)]

= A exp[-i/ Ћ(Et-px)](i/ Ћ p)

dΨ/ dx = Aip/ Ћ exp[-i/ Ћ(Et-px)]

Again differentiating wr.t. x

dΨ/d x² =Aip/  Ћ exp[ -i/ Ћ(Et-px)](i/ Ћ p)

= A(ip/ Ћ)² exp [-i/ Ћ(Et-px)]

dΨ/dx²= -p²/h² Ψ                                       (3)[ using equation(2)]

Or                     p² Ψ= – Ћ² d²Ψ/ dx²

Also on differentiating (2) wr.t. t ,we get

dΨ(x,t)/dt= A exp [-i/ Ћ(Et-px)] (-i/ Ћ E)

dΨ/dt =-(i/ Ћ) E Ψ

or                  E Ψ=(i Ћ) Ψ/ t                                                   (4)

When the particle is acted upon by a force then its total energy is the sum of Kinetic and potential energies i.e.

Total energy = Kinetic energy + potential energy

E=p²/2m +V

E Ψ= p²/2m Ψ+ V Ψ                                                        (5)

Putting equations (3) and (4) in equations (5), we get

-Ћ/t dΨ/dt=-h²/2m d²Ψ/dx²+V Ψ

i Ћ dΨ/dt= – Ћ²/2m( Ψ/x²) + V Ψ                           (6)

Which is time dependent form of Schroedinger wave equation

In three –dimensional form

i Ћ Ψ/t =- Ћ ²/2m(d²Ψ/dx²+ d²Ψ/dy²+ dΨ/dz²)+ V Ψ

where the particle potential V is a function of x,y,z and t . Any restriction on the particle motion will effect the potential energy V. once V is known, Schroedinger equation may be solved for the wave function Ψ of the particle form where Ψ² may be determined.

The post TIME DEPENDENT SCHRODINGER WAVE EQUATION first appeared on Winner Science.

Following examples will show that classical mechanics was inadequate to give explanation of observed facts:

(a) Photoelectric effect: when a photon of light falls on the metal surface and if the frequency of light is more than the certain minimum frequency (called threshold frequency), then electrons are ejected from metal surface. This effect is known as photoelectric effect.

According to classical theory of light, there should be no threshold frequency, the photoelectric current should increase with increase in frequency of light and kinetic energy of ejected electrons should increase with increase in intensity of light. There must also be a finite delay between the incidence of photons and ejection of electrons from metal surface.

But experimental results were totally opposite ,like, photoelectric effect is instantaneous ,and other observations were contradicted to aforementioned classical theory based observations.

Therefore classical mechanics failed to explain the photoelectric effect. These observations successfully later on by Einstein’s theory of photoelectric emission, which is based on Planck’s Quantum theory of Radiation.

(b) Stability of atom: Atomic model prepared by Rutherford was based on classical physics. This model assumes the atom to be consisting of a positively charged nucleus in the centre and negatively charged electrons revolved the nucleus in the circular orbit. Classical theory says that whenever a charged particle undergoes accelerated motion, it emits electromagnetic radiation. Therefore, an electron must emit energy continuously as its motion is accelerated. Thus, the orbital radius should decrease continuously and ultimately electron would fall in to nucleus. Therefore, an atom would collapse at the end.

But, this does not happen.   As atom is stable and stability of atom could not be established by classical physics .This stability of atom was explained by Bohr’s quantum theory of atom.

Conclusion

Similarly, there were a lot of other observations like black body radiation, optical spectra, Compton Effect etc. which were not explained by classical mechanics. Therefore, it can be concluded that classical mechanics is inadequate to discuss the subatomic phenomenon. Hence, there is a need to introduce a new theory, which can deal with mechanics or motion of the subatomic or microscopic particles. Thus, the quantum hypothesis originated and the branch of physics dealing with motion of subatomic particles is called quantum mechanics.

The post Origin of Quantum Physics first appeared on Winner Science.

If ∆x and ∆p_x are the uncertainties in the simultaneous measurements of position and momentum of the electron in the first orbit, then from uncertainty principle

∆x∆p_x = Ћ

Where Ћ = h/2∏

Or    ∆p_x = Ћ /∆x                                                                 (1)

As kinetic energy is given as

K = p²/2m

Then uncertainty in K.E is

∆K =∆p²_x/2m

Put equation (i) in above equation

∆K= Ћ² /2m(∆x)² (2)

As potential energy is given by

∆V= -1/4∏ε₀ Ze²/∆x                                                      (3)

The uncertainty in total energy is given by adding equations (2) and (3), that is

∆E= ∆K+∆V

= Ћ² /2∏(∆x)² –Ze²/4∏ε₀∆x

If ∆x = r= radius of Bohr’s orbit, then

∆E= Ћ² /2mr² –Ze²/4∏ε₀r                                                   (4)

The Uncertainty in total energy will be minimum if

d(∆E)/dr=0 and d²⁽(∆E)/dr² is positive

Differentiating equation (4) w.r.t. r, we get

d(∆E)/dr=0= – Ћ ²/mr³+Ze²/4π ε₀r² (5)

For minimum value of ∆E

d(∆E)/dr=0= – Ћ ²/mr²+Ze²/4π ε₀r²

or                  Ze²/4π ε₀r²= Ћ ²/mr³

Or                       r=4π ε₀ Ћ ²/me² (6)

Further differentiating equation (5), we get

d²(∆E)/dr²=3 Ћ ²/mr⁴-2Ze²/4π ε₀r³

By putting value of r from equation (6) in above equation, we get positive value of

d²(∆E)/dr²

Therefore equation (4) represents the condition of minimum in the first orbit.

Hence, the radius of first orbit is given by

r=4π ε₀Ћ ²/me²=0.53 angstrom                                  (For H atom Z=1)

Put value of r in equation (4), we get

E_min= -13.6 e V

This value is same as determined by using Bohr’s theory.

Therefore, with the help of Heisenberg’s uncertainty principle, one can determine the radius of the Bohr’s first orbit.

The post Applications of the Heisenberg Uncertainty Principle: The Radius of Bohr’s First Orbit first appeared on Winner Science.

In 1924, Lewis de-Broglie proposed that matter has dual characteristic just like radiation. His concept about the dual nature of matter was based on the following observations:-

(a)    The whole universe is composed of matter and electromagnetic radiations. Since both are forms of energy so can be transformed into each other.

(b)   The matter loves symmetry.  As the radiation has dual nature, matter should also possess dual character.

According to the de Broglie concept of matter waves, the matter has dual nature. It means when the matter is moving it shows the wave properties (like interference, diffraction etc.) are associated with it and when it is in the state of rest then it shows particle properties. Thus the matter has dual nature. The waves associated with moving particles are matter waves or de-Broglie waves.

WAVELENGTH OF DE-BROGLIE WAVES

Consider a photon whose energy is given by

E=hυ=hc/λ            – – (1)

If a photon possesses mass (rest mass is zero), then according to the theory of relatively ,its energy is given by

E=mc² – – (2)

From (1) and (2) ,we have

Mass of photon m= h/cλ

Therefore Momentum of photon

P=mc=hc/cλ=h/λ    – – (3)

Or           λ = h/p

If instead of a photon,  we consider a material particle of mass m moving with velocity v,then the momentum of the particle ,p=mv. Therefore, the wavelength of the wave associated with this moving particle is given by:

h/mv                                      –

Or           λ = h/p  (But here p = mv)           (4)

This wavelength is called DE-Broglie wavelength.

Special Cases:

1. dE-Broglie wavelength for material particle:

If E is the kinetic energy of the material particle of mass m moving with velocity v,then

E=1/2 mv²=1/2 m²v²=p²/2m

Or              p=√2mE

Therefore the by putting above equation in equation (4), we get de-Broglie wavelength equation for material particle as:

λ = h/√2mE     – – (5)

2. dE-Broglie wavelength for particle in gaseous state:

According to kinetic theory of gases , the average kinetic energy of the material particle is given by

E=(3/2) kT

Where k=1.38 x 10^-23 J/K is the Boltzmann’s constant  and T is the absolute temperature of the particle.

Also E = p²/2m

Comparing above two equations, we get:

p²/2m = (3/2) kT

or p = /√3mKT

Therefore   Equation (4) becomes

λ=h/√3mKT

This is the dE-Broglie wavelength for particle in gaseous state:

3. dE-Broglie wavelength for an accelerated electron:

Suppose an electron accelerates through a potential difference of V volt. The work done by electric field on the electron appears as the gain in its kinetic energy

That is E = eV

Also E = p²/2m

Where e is the charge on the electron, m is the mass of electron and v is the velocity of electron, then

Comparing above two equations, we get:

eV= p²/2m

or  p = √2meV

Thus by putting this equation in equation (4), we get the the de-Broglie wavelength of the electron as

λ  = h/√2meV  6.63 x 10^-34/√2 x 9.1 x 10-³¹ x1.6 x 10^-19 V

λ=12.27/√V  Å

This is the de-Broglie wavelength for electron moving in a potential difference of V volt.

Brain Teaser:

If you have understood the article, then please answer the following question:

1. If an electron is accelerated under the potential of 100 volts, then what should be the wavelength of electron?

a) 12.27 Å

b) 1.27 Å

c) 10 Å

d) 100 Å

Please submit your answer in comments section.

The post de Broglie concept of matter waves: dual nature of matter first appeared on Winner Science.

The Heisenberg uncertainty principle based on quantum physics explains a number of facts which could not be explained by classical physics. One of the applications is to prove that electron can not exist inside the nucleus. It is as follows:-

Non-existence of electrons in the nucleus

In this article, we will prove that electrons cannot exist inside the nucleus.

But to prove it, let us assume that electrons exist in the nucleus. As the radius of the nucleus in approximately 10^-14 m. If electron is to exist inside the nucleus, then uncertainty in the position of the electron is given by

∆x= 10^-14 m

According to uncertainty principle,

∆x∆p_x =h/2∏

Thus                            ∆p_x=h/2∏∆x

Or                               ∆p_x =6.62 x10^-34/2 x 3.14 x 10^-14

∆p_x=1.05 x 10^-20 kg m/ sec

If this is p the uncertainty in the momentum of electron ,then the momentum of electron should be at least of this order, that is p=1.05*10^-20 kg m/sec.

An electron having this much high momentum must have a velocity comparable to the velocity of light. Thus, its energy should be calculated by the following relativistic formula

E=  √ m²₀ c⁴ + p²c²

E =  √(9.1*10^-31)² (3*10⁸)⁴ + (1.05*10^-20)²(3*10⁸)²

= √(6707.61*10^-30) +(9.92*10^-24)

=(0.006707*10^-24) +(9.92*10^-24)

= √9.9267*10^-24

E= 3.15*10^-12 J

Or                                  E=3.15*10^-12/1.6*10^-19 eV

E= 19.6* 10⁶ eV

Or                                  E= 19.6 MeV

Therefore, if the electron exists in the nucleus, it should have an energy of the order of 19.6 MeV. However, it is observed that beta-particles (electrons) ejected from the nucleus during b –decay have energies of approximately 3 Me V, which is quite different from the calculated value of 19.6 MeV. Second reason that electron can not exist inside the nucleus is that experimental results show that no electron or particle in the atom possess energy greater than 4 MeV.

Therefore, it is confirmed that electrons do not exist inside the nucleus.

The post Applications of Heisenberg’s Uncertainty principle: Non-existence of electrons in the nucleus first appeared on Winner Science.

We know, group velocity of the wave packet is given by

v_g =∆ω/∆k

Where ω is the angular frequency and k is the propagation constant or wave number

But v_g is equal to the particle velocity v

Thus v_g = v =  ∆ω/∆k                                                   (1)

But                    ω=2пf

Where f is the frequency

Therefore   ∆ ω = 2п ∆ f                          (2)

Also                       k=2 п/λ

Since         de-Broglie wavelength   λ=h/p

By putting this value in equation of k, we get

k=2пp/ λ

Therefore                ∆k=2п∆p / λ                                 (3)

Put equations (2) and (3) in equation (1), we get

v= 2пh∆f/2п∆p =h∆f /            (4)

Let the particle covers distance ∆x in time ∆t, then particle velocity is given by

v  = ∆x/∆t                     (5)

Compare equations (4) and (5), we get

∆x/∆t=h∆f/∆p

Or                          ∆x.∆p=h∆f ∆t                                     (6)

The frequency ∆f is related to ∆t by relation

∆t≥ 1/∆f                                           (7)

Hence equations (6) becomes

∆x.∆p≥ h

A more sophisticated derivation of Heisenberg’s uncertainty principle  gives

∆x.∆p=h/2п                                          (8)

Which is the expression of the Heisenberg uncertainty principle.

As the particle is moving along x-axis. Therefore, the momentum in equation (8) of Heisenberg’s uncertainty principle should be the component of the momentum in the x-direction, thus equation Heisenberg’s uncertainty principle can be written as,

∆x.∆p_x=h/2п                             (9)

Note: There can not be any uncertainty if momentum is along y direction.

Q: Why there is uncertainty in position and momentum?

Answer: Because the particle is always in disturbed state during motion. It is not possible to calculate the position and momentum of particle simultaneously.

The post Heisenberg uncertainty principle first appeared on Winner Science.