De broglie wavelength equation (De-Broglie equation )

De Broglie Wavelength Formula

        In some situations, light behaves like a wave, while in others, it behaves like particles. The particles of light are called photons, and they can be thought of as both waves and particles. Louis de Broglie (1892-1987) developed a formula to relate this dual wave and particle behavior. It can also be applied to other particles, like electrons and protons. The formula relates the wavelength to the momentum of a wave/particle.

For particles with mass (electrons, protons, etc., but not photons), there is another form of the de Broglie wavelength formula. At non-relativistic speeds, the momentum of a particle is equal to its rest mass, m, multiplied by its velocity, v.

The unit of the de Broglie wavelength is meters (m), though it is often very small, and so expressed in nanometers (1 nm = 10(-9) m), or Angstroms  

 De broglie wavelength equation (De-Broglie equation )

     In 1924, Louis de-Broglie give the idea that matter should also posses dual nature. According to Louis de-Broglie a moving matter particle is surrounded by a wave whose wavelengths depend upon the mass of the particle and its velocity. These waves associated with the matter particles are known as matter waves or de-Broglie waves. The wavelength of the particle can be find through the following relation - 

                  

where h is the Planck's constant and p is the momentum of particle. The value of Planck's constant is 

 The de-Broglie concept of matter waves was based on the following facts- 

(i) Matter and light, both are forms of energy and each of them can be transformed into the other.

(ii) Both are governed by the space time symmetries of the theory of relativity.

Derive an expression for the de broglie wavelength:-

Considering the Planck's theory of radiation, the energy of the a photon (quantum) is given by

                              ......(1)

where c is the  velocity of light in vacuum and  is its wavelength. 

According to Einstein energy mass relation 

                              ........(2)

From equation (1) & (2), we get 

             ........(3)

where mc = p (momentum associated with photon).

If we consider the case of a material particle of mass m and moving with a velocity v, then the wavelength associated with this particle is given by 

                       ..........(4)

If E is the kinetic energy of the material particle then 

                      

            .........(5)

Substituting the value of p in equation (4), we get  de broglie wavelength,

                ...........(6)

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

where, K=  Boltzmann's constant = 

Putting the value of E in equation (4)

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Home rothery's rules and regulations (physics)

Hume-Rothery rules:-

HUME-ROTHERY'S RULES:- 

             In the course of an alloy development, it is frequently desirable to increase the strength of the alloy by adding a metal that will form a solid solution. In the choice of such alloying elements, a number of rules govern the for- mation of substitutional work of Hume-Rothery. Unfortunately, if an alloying element is chosen at random, intermediate phase instead of a solid solution. 

Hume-Rothery's Rules are described below. is likely to form an objectionable.

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CHEMICAL AFFINITY FACTOR 

             The greater the chemical affinity of two metals the more restricted is their solid solubility. When their chemical affinity is great, two metals tend to form an intermediate phase rather than a solid solution.

             Chemical Affinity Factor-The greater the chemical affinity of two metals, the more restricted is their solid solubility. When their chemical affinity is great, two metals tend to form an intermediate phase rather than a solid-solution. Generally, the farther apart the elements are in the periodic table, the greater is their chemical affinity.

RELATIVE VALENCY FACTOR 

              If the alloying element has a different valence from that of the base metal, the number of valence electrons per atom, called the electron ratio, will be changed by alloying. Crystal structures are more sensitive to a decrease in the electron ratio than to an increase. Therefore, a metal of high valence can dissolve only a small amount of lower valance metal, while the lower valence metal may have good solu- bility for a higher valence metal. 

              Relative Valence (Valency) Factor - It has been found that the metal of high valence can dissolve only a small amount of a lower valence metal, while the lower valence metal may have good solubility for the higher valence metal. For example in the Al-Ni alloy system both metals have FCC structure. The relative size factor is approximately 14%. However, Ni is lower n valence than Al and thus solid nickel dissolves 5% aluminium, but the higher valence Al dissolves only 0.04% Ni.

RELATIVE SIZE FACTOR 

            If the size of two metallic atoms (given approxima- tely by their constants) differs by less than 15 percent, the metals are said to have a favourable size factor for solid solution formation. So far as this factor is concerned, each of the metals will be able to dissolve apprecia- bly (to the order of l0%) in the other metal. If the size factor is greater than 15°, solid solution formation tends to be severely limited and is usually only a fraction of one percent.

               Relative Size Factor- If the two metals are to exhibit extensive solid solubility in each other, it is essential, that their atomic diameters shall be fairly similar, since atoms differing greatly in size cannot be accommodated readily in the same structure (as a substitutional solid solution) without producing excessive strain and corresponding instability. This is what referred as the term size factor. The extensive solid solubility is encountered only when the two different atoms differ in size by less than 15%, calleda favourable size factor (e.g. Cu-Ni, Au-Pt). If the relative size factor is between 8% and 15 % , the alloy system usually shows a minimum and if this is greater than 15 % , substitutional solid solution formation is very limited. 

LATTICE-TYPE FACTOR 

           Only metals that have the same type of lattice (FCC for example) can form a complete series of solid solutions. Also, for com- plete solid solubility, the size factor must usually be less than 8 percent. Copper-nickel and silver-gold-platinum are examples of binary and ter- nary systems exhibiting complete solid solubility.

              Crystal Structure Factor The crystal lattice structure of the two elements (metal) should be same (i.e. both should be of BCC, FCC, or HCP structure) for complete solubility, otherwise the two solutions would not merge into each other. Also for complete solid solubility the size must usually be less than 8%.

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Hume-Rothery's Rules are described below. is likely to form an objectionable.

रासायनिक प्रभाव क्षेत्र

             दो धातुओं का रासायनिक आत्मीयता जितना अधिक प्रतिबंधित है, उनकी ठोस घुलनशीलता है। जब उनकी रासायनिक आत्मीयता महान होती है, तो दो धातुएं एक ठोस समाधान के बजाय एक मध्यवर्ती चरण बनाती हैं।

रिश्ता वैधता का कारखाना

              यदि मिश्र धातु तत्व का आधार धातु से भिन्न भिन्नता है, तो प्रति परमाणुओं में वैलेंस इलेक्ट्रॉनों की संख्या, जिसे इलेक्ट्रॉन अनुपात कहा जाता है, को मिश्रधातु द्वारा बदल दिया जाएगा। वृद्धि की तुलना में क्रिस्टल संरचनाएं इलेक्ट्रॉन अनुपात में कमी के प्रति अधिक संवेदनशील हैं। इसलिए, उच्च वैलेंस की एक धातु केवल कम वैलेंस धातु को भंग कर सकती है, जबकि कम वैलेंस धातु में उच्च वैलेंस धातु के लिए अच्छा सॉल्यू-बाइट हो सकता है।

विश्वसनीय आकार फैक्टरी

            यदि दो धात्विक परमाणुओं का आकार (उनके स्थिरांक द्वारा अनुमानित रूप से दिया गया) 15 प्रतिशत से कम होता है, तो धातु को ठोस घोल बनाने के लिए अनुकूल आकार कारक कहा जाता है। जहाँ तक इस कारक का संबंध है, प्रत्येक धातु दूसरे धातु में apprecia- bly (l0% के क्रम में) को भंग करने में सक्षम होगी। यदि आकार का कारक 15 ° से अधिक है, तो ठोस समाधान गठन गंभीर रूप से सीमित हो जाता है और आमतौर पर केवल एक प्रतिशत का एक अंश होता है।

लेटिस-टाईप फैक्टर

           केवल धातुएँ जिनमें एक ही प्रकार की जाली होती है (उदाहरण के लिए FCC) ठोस समाधानों की एक पूरी श्रृंखला बना सकती है। इसके अलावा, कॉम-पिल सॉलिड सॉल्युबिलिटी के लिए, साइज फैक्टर आमतौर पर 8 प्रतिशत से कम होना चाहिए। कॉपर-निकेल और सिल्वर-गोल्ड-प्लैटिनम बाइनरी और टेर-नैरी सिस्टम के उदाहरण हैं जो पूर्ण ठोस घुलनशीलता प्रदर्शित करते हैं।

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CONTINUOUS CASTING PROCESS (Asarco Process)

Continuous casting process:-

In Continuous casting process the forming die is an integral part of the furnace and thus there is no problem of controlling the flow of metal . The metal is fed by gravity into the mould from the furnace as it is continuously solidified and withdrawn by the rolls below . The die is water cooled and self lubricating and thus has excellent resistance to thermal shocks . The upper end of the die is in molten metal and thus serves the function of riser and acts as path for dissipation of evolved gases .
      At the time of starting , a similar rod is put into the die till its upper end melts and forms Holding continuity with molten metal , and it is then Furnace withdrawn by rolls and the process becomes continuous . By this process , rounds , tubes , squares and special shapes can be conveniently produced and it is particularly suited to alloys like phosphorised copper and standard bronzes .

      The physical properties of castings Casting - produced are superior to other processes . In continuous casting , solidifying zone being relatively small , practically all problems encountered with feeding and shrinkage are overcome . Problems due to fast cooling in mould zone are faced segregation and cracking ) . A greater degree of control is required in comparison batch casting . Asarco process is shown n fig.

 Uses - Continuous casting process is specially suited to rounds , tubes , squares and special shapes , particularly made from alloys like phosphorized popper and standard bronzes .

Merits - ( i ) Physical properties of castings produced are superior to other processes , because metal is protected from contamination , while melting and being poured .

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Examples of differential equations with solutions

“Ordinary Differential Equations:-

           In mathematics, the term “Ordinary Differential Equations” also known as ODE is a relation that contains only one independent variable and one or more of its derivatives with respect to the variable. In other words, the ODE’S is represented as the relation having one real variable x, the real dependent variable y, with some of its derivatives.

Y’,y”, ….yn ,…with respect to x.

           The order of ordinary differential equations is defined to be the order of the highest derivative that occurs in the equation. The general form of n-th order ODE is given as;

F(x, y,y’,….,yn ) = 0

Note that, y’ can be either dy/dx or dy/dt and yn can be either dny/dxn or dny/dtn.

An n-th order ordinary differential equations is linear if it can be written in the form;

a0(x)yn + a1(x)yn-1 +…..+ an(x)y = r(x)

          (The function aj(x), 0≤j≤n are called the coefficients of the linear equation. The equation is said to be homogeneous if r(x)=0.If r(x)≠0 , it is said to be a non- homogeneous equation. Also, learn the first order differential equation here.)

Ordinary Differential Equations Types

              The ordinary differential equation is further classified into three types. They are:

• Autonomous ODE
• Linear ODE
• Non-linear ODE

Autonomous Ordinary Differential Equations

A differential equation which does not depend on the variable, say x is known as an autonomous differential equation.

Linear Ordinary Differential Equations

       If differential equations can be written as the linear combinations of the derivatives of y, then it is known as linear ordinary differential equations. It is further classified into two types,

• Homogeneous linear differential equations
• Non-homogeneous linear differential equations

Non-linear Ordinary Differential Equations

If differential equations cannot be written in the form of linear combinations of the derivatives of y, then it is known as non-linear ordinary differential equations.

Ordinary Differential Equations Application

ODEs has remarkable applications and it has the ability to predict the world around us. It is used in a variety of disciplines like biology, economics, physics, chemistry and engineering. It helps to predict the exponential growth and decay, population and species growth. Some of the uses of ODEs are:

• Modelling the growth of diseases
• Describes the movement of electricity
• Describes the motion of the pendulum, waves
• Used in Newton’s second law of motion and Law of cooling.

Ordinary Differential Equations Examples

Some of the examples of ODEs are as follows;

Ordinary Differential Equations Problems and Solutions

          The ordinary differential equations solutions are found in an easy way with the help of integration. Go through once and get the knowledge of how to solve the problem.   

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Question 1:

Find the solution to the ordinary differential equation y’=2x+1

Solution:

Given, y’=2x+1

Now integrate on both sides, ∫ y’dx = ∫ (2x+1)dx

Which gives, y=2x2/2+x+c

y=x2+x+c

Where c is an arbitrary constant.

Question 2:

Solve y4y’+ y’+ x2 + 1 = 0

Solution:

Take, y’ as common,

y'(y4+1)=-x2-1

Now integrate on both sides, we get

y55+y=−x33−x+c

Where c is an arbitrary constant.

     For more maths concepts, keep visiting BYJU’S and get various maths related videos to understand the concept in an easy and engaging way.

Imp.Q.

        Solve                Given that  is a solution.

sol.

     The given equation can be written in the standard form as 

                              ........(1)

Here 1+P+Q=0, therefore   is a part of the C.F. of the solution of equation (1).

putting    & the corresponding values of  in equation (1), we get

     or                         where        

  or              (on integration)

   or     

   The complete salution of equation (1) is              👈  Ans

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Thermal stresses induced in bars of tapering section due to change in temperature_

Define Temperature Stresses :-


                 When ever there is some increase or decrease in temperature of a body, it cause the body to expand or contract. If the boys is allowed to expand or contract freely, with the rise or fall of the temperature, no stresses will be induced in the body. But if the deformation of the body is prevented some stresses are induced in the body. Such stresses are called thermal or temperature stresses.
                 The corresponding strains are called thermal & temperature strain.

                 Temperature stresses in beams or rods can be calculated as follows -
(i) Determine its expansion or contraction due to change in temperature assuming it is free to expand or contract.
(ii) Calculate the load required to bring the beam in original position.
(iii) Calculate stress and strain corresponding to this load.

 l = Original length of the beam
t1 = Initial temperature of the beam
t2 = Final temperature of the beam
 = Coefficient of linear expansion of beam material.
          (Elongation of the beam due to increase in temperature 
 If elongation of the beam is prevented by some external force or by fixing its end, temperature strain will produced in beam, which is given by,

                                

and temperature stress 




Q. A bar shown in fig. is subjected to axial forces and fixed at L and P. determine the forces in each portion of the bar and displacements of point M and N. Take  

         n
Sol     =>        

          Given,    




Forces in each portion 

   As we can see from figure, portion LM will be in tension, while portion NP will be in compression.
   The free body diagram of all three portion are shown in fig.
Now, for static equilibrium of the bar,

   
     R1+R2=100+50=150kN                       ........(1)
As bar is fixed at the ends, therefor extension of portion of portion LM will be equal to the compressions of MN & NP i.e,.
                                      

           

 

                 

                     

                   ............(2)

On solving equations (1)&(2), we get

 & 

Hence, forces in each portion,

      66.67 kN (tensile)                 

 

      

      

                 16.67 kN (compressive)              Ans

& 83.33 kN (compressive )              Ans

Displacement of point M,

 0.1587 mm    Ans

Displacement of point N,

here      

    Displacement of point N,

     0.1984 mm                    Ans

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