De Broglie Equation For Velocity

De-Broglie Wavelength Formula: Equations, Solved Examples

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De-Broglie Wavelength Formula – Einstein proposed that any electromagnetic radiation, including light which was, till then, considered an electromagnetic wave, in fact, showed particle-like nature. He coined the word "photon" for the quanta or particle of light. Soon, scientists began to wonder if other particles could also have a dual wave-particle nature.

In 1924, French scientist Louis de Broglie derived an equation, known as the De Broglie Wavelength Formula, that described the moving ridge nature of whatever particle. Thus, establishing the wave-particle duality for the matter. Microscopic particle-like electrons likewise proved to possess this dual nature holding. Permit us learn about the equation proposed by de-Broglie in detail in this article.

De Broglie's Hypothesis and Equation

Louis-de-Broglie explained the concept of de-Broglie waves in the twelvemonth 1923. In his thesis, he suggested that any moving particle, whether microscopic or macroscopic, volition be related to a wave graphic symbol. This was later experimented with and proved by Davisson and Germer inside the year 1927. The waves associated with thing were called 'Matter Waves'. These waves explain the character of the wave associated with the particle. We know that electromagnetic radiation exhibit the dual nature of a particle (having a momentum) and wave (expressed in frequency, and wavelength).

He further proposed a relation between the speed and momentum with the wavelength if the particle had to behave as a wave. Since, at not-relativistic speeds, the momentum of a particle will be adequate to its mass \(\text {thousand}\), multiplied by its velocity \(\text {five}\). Thus, according to de-Broglie, the wavelength \(\left( \lambda \correct)\) of whatsoever moving object is given past:

\(\lambda = \frac{{\rm{h}}}{{\rm{p}}}\)

Where \(\text {h}\) is Planck's constant and \(\text {p}\) is the momentum of the particle.

Derivation of De Broglie'southward Wavelength

From Einstein'southward relation of mass-energy equivalence, we know that,

\({\text{E}} = {\text{m}}{{\text{c}}^2} \cdots (ane)\)

Where,

\(\text {East}=\) energy of the particle

\(\text {1000}=\) mass of the particle

\(\text {c}=\) speed of lite

According to Planck'south theory, every breakthrough of a wave has a discrete amount of energy associated with information technology, and he gave the equation:

\({\text{E}} = {\text{hf}} \cdots (2)\)

Where,

\(\text {E}:\) energy of the particle

\(\text {h}=half dozen.62607 \times 10^{-34} \mathrm{Js}:\) Planck's constant

\(\text {f}=\) frequency

De-Broglie's hypothesis suggested that particles and waves deport as similar entities. Thus, he equated the free energy relation for both particle and wave; equating equations \((i)\) and \((2)\), we go:

\(\text {mc}^{2}=\text {hf}\)

Since the particles more often than not exercise not travel at the speed of calorie-free, De Broglie substituted the speed of light \(\text {c}\), with the velocity of a real particle \(\text {v}\), and obtained:

\({\text{m}}{{\text{v}}^2} = {\text{hf}} \cdots (3)\)

If \(\lambda \) exist the wavelength of the moving ridge, then the frequency will be: \({\rm{f}} = \frac{{\rm{v}}}{\lambda }\)

Substituting this in equation \((3)\), we become:

\({\rm{m}}{{\rm{v}}^2} = \frac{{{\rm{hv}}}}{\lambda }\)

\(\lambda = \frac{{\rm{h}}}{{{\rm{mv}}}}\)

or, \(\lambda = \frac{{\rm{h}}}{{\rm{p}}}\,\,\, \cdots (iv)\)

Where \(\text {p}\) is the momentum of the particle.

De Broglie Wavelength and Kinetic Energy

The kinetic energy of an object of mass \(\text {m}\) moving with velocity \(\text {five}\) is given as:

\({\text{G}} = \frac{1}{2}{\text{m}}{{\text{v}}^2}\)

or, \({\text{Yard}} = \frac{1}{2}{\text{mv}} \cdot {\text{five}}\)

\({\text{m}} \cdot {\text{G}} = \frac{one}{2}{({\text{mv}})^2}\)

Since, \(\text {p}=\text {mv}\), Thus:

\({\text{m}}.{\text{K}} = \frac{i}{2}{({\text{p}})^2}\)

From equation \((4),\,{\rm{p}} = \frac{{\rm{h}}}{\lambda }\)

\( \Rightarrow {\rm{m}} \cdot {\rm{K}} = \frac{1}{ii}{\left( {\frac{{\rm{h}}}{\lambda }} \right)^2}\)

\({\lambda ^ii} = \frac{{{{\rm{h}}^2}}}{{2\,{\rm{mK}}}}\)

\(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{mK}}} }}\)

De-Broglie Wavelength and Potential

When a charged particle, having a charge \(\text {q}\) is accelerated through an external potential difference \(\text {V}\), the free energy of the particle can exist given equally:

\({\text{E}} = {\text{qV}} \cdots ({\text{i}})\)

According to Planck'southward equation,

\(\text {E}=\text {hf}\)

Since, \({\rm{f}} = \frac{{\rm{v}}}{\lambda }\)

Therefore, \({\rm{E}} = {\rm{h}}\frac{{\rm{five}}}{\lambda }\,\,\, \ldots ({\rm{ii}})\)

Equating the equations \(\left({\text{i}} \right)\) and \(\left({\text{ii}} \right)\),

\({\rm{qV}} = {\rm{h}}\frac{{\rm{v}}}{\lambda }\)

or, \(\lambda = \frac{{{\rm{hv}}}}{{{\rm{qV}}}}\)

Thermal De Broglie Wavelength

In that location exists a relation betwixt the De-Broglie equation and the temperature of the given gas molecules, and the thermal de Broglie wavelength gives information technology \(\left( {{\lambda _{{\rm{Th}}}}} \right).\) The Thermal de Broglie equation represents the average value of the de Broglie wavelength of the gas particles at the specified temperature in an platonic gas.

The expression gives the thermal de Broglie wavelength at temperature \(\text {T}\):

\({\lambda _{{\rm{Thursday}}}} = \lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{thou}}{{\rm{g}}_{\rm{B}}}{\rm{T}}} }}\)

where,

\(h=\) Planck constant

\(thou=\) mass of a gas particle

\({{\text{k}}_{\text{B}}} = \) Boltzmann constant

De Broglie Wavelength of Twenty-four hours-to-Day Objects

According to the hypothesis, all particles accept a moving ridge associated with them. That is truthful for us humans and the objects around us. To get an idea of the de-Broglie wavelength associated with macroscopic particles:

Let us detect the wavelength of a wave associated with a car of mass \(1000 \mathrm{~kg}\) moving with the velocity of \(10 \mathrm{~m} / \mathrm{southward}.\)

De Broglie Wavelength of Day-to-Day Objects

The wavelength associated with the car will be: \(\lambda = \frac{{\rm{h}}}{{{\rm{mv}}}}\)

\(\lambda=\frac{6.62607 \times 10^{-34} \mathrm{Js}}{grand \mathrm{~kg} \times x \mathrm{~m} / \mathrm{south}}=vi.6 \times 10^{-30} \mathrm{~m}=6.half-dozen \times 10^{-21} \mathrm{~nm}\)

Thus, the value of wavelength associated with this automobile is insignificant.

Similarly, for other macroscopic objects with big mass values, the wavelength associated with them is so small that it can not be detected.

De-Broglie Wavelength of an Electron

Every bit we have seen in a higher place, the matter waves associated with real objects is so minor that it is of no skillful use to u.s.a.. But for sub-atomic particles with negligible masses, the value of de-Broglie wavelength is substantial. To calculate the de-Broglie wavelength associated with a microscopic particle,

Permit us take an electron of mass \({\rm{m}} = 9.1 \times {10^{ – 31}}\;{\rm{kg}},\) moving with the speed of light, i.e., \(\text {c}=3 \times 10^{8} \mathrm{~1000} / \mathrm{due south}\), then the de-Broglie wavelength associated with information technology can exist given as:

\(\lambda = \frac{{\rm{h}}}{{{\rm{mc}}}}\)

\(\lambda=\frac{half dozen.62607 \times x^{-34} \mathrm{Js}}{nine.one \times x^{-31} \mathrm{~kg} \times 3 \times 10^{8} \mathrm{~m} / \mathrm{south}}=0.7318 \times 10^{-11} \mathrm{~m}=0.073 \mathrm{~A}^{\circ}\)

This is a substantial value. Thus, the de-Broglie wavelength associated has a pregnant value, and it can be detected.

Relation Between De-Broglie Wavelength and Potential (for an electron)

The expression for the de-Broglie wavelength of an electron,

\(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{mK}}} }}\)

If the electron having a charge e is moving under an external potential \(\text {V}\), and so,

The kinetic energy of the electron, \({\text{M}} = {\text{eV}}\)

Substituting this expression in the above equation,

\(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{meV}}} }}\)

Put, \(h=6.62607 \times 10^{-34} \mathrm{Js}\)

\(\text {e}=i.6 \times 10^{-19} \mathrm{C}\)

\(\text {m}=9.i \times 10^{-31} \mathrm{~kg}\)

\(\lambda = \frac{{12.27}}{{\sqrt {\rm{V}} }}{\rm{A}}^\circ \)

De Broglie Wavelength- Solved Problems

Q.one. A certain photon has a momentum of \(i.50 \times x^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}\). What will be the photon's de Broglie wavelength?

Ans: \(\text {p}=one.fifty \times 10^{-27} \mathrm{~kg} \mathrm{~thou} / \mathrm{s}\)
Plank's abiding, \(\text {h}=half-dozen.62607 \times 10^{-34} \mathrm{Js}\)
The de Broglie wavelength of the photon can be computed using the formula:
\(\lambda = \frac{{\rm{h}}}{{\rm{p}}}\)
\(=\frac{6.62607 \times ten^{-34} \mathrm{Js}}{1.fifty \times 10^{-27} \mathrm{kgm} / \mathrm{due south}}\)
\(=4.42 \times x^{-7} \mathrm{~m}\)
\(=442 \times x^{-9} \mathrm{~yard}\)
\(=442 \mathrm{~nm}\)
The de Broglie wavelength of the photon volition exist \(442 \mathrm{~nm}\), and this wavelength lies in the blue-violet part of the visible light spectrum.

Q.2. What is the de Broglie wavelength of an electron which is accelerated through a potential difference of \(10\, \mathrm{kV}\) ?

Ans: If the electron having a charge \(\text {e}\) is moving nether an external potential \(\text {Five}\), and so the expression for the de-Broglie wavelength of an electron is:
\(\lambda = \frac{{12.27}}{{\sqrt {\rm{V}} }}{\rm{A}}^\circ \)
Nosotros are given, \(\text {V}=x \mathrm{kV}=10 \times 10^{3} \mathrm{~V}=ten^{four} \mathrm{~V}\)
\(\lambda = \frac{{12.27}}{{\sqrt {{{10}^4}} }}{\rm{A}}^\circ \)
\(\lambda = \frac{{12.27}}{{100}}{\rm{A}}^\circ = 0.1227{\mkern 1mu} {\rm{A}}^\circ \)

Summary

Co-ordinate to de-Broglie, the wavelength \(\left( \lambda \right)\) of whatever moving object is given past: \(\lambda = \frac{{\rm{h}}}{{\rm{p}}},\) Where \(\text {h}\) is Planck'south constant and \(\text {p}\) is the mass of the particle.

The relation betwixt de-Broglie wavelength and the kinetic energy of an object of mass \(\text {m}\) moving with velocity \(\text {v}\) is given as: \(\lambda = \frac{{\rm{h}}}{{\sqrt {ii\,{\rm{mK}}} }}\)

When a charged particle having a charge \(\text {q}\) is accelerated through an external potential departure \(\text {V}\), de-Broglie wavelength, \(\lambda = \frac{{{\rm{hv}}}}{{{\rm{qV}}}}\)

The expression for the de-Broglie wavelength of an electron, \(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{mK}}} }}\) or \(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{meV}}} }}\)

Frequently Asked Questions on De Broglie Wavelength Formula

The virtually unremarkably asked questions on De Broglie Wavelength Formula are answered here:

Q.1. For the same value of de Broglie wavelength, which has greater speed: Electron or Proton?
Ans: The de-Broglie wavelength of the particle is the same. Thus their momentum will be equal. Momentum is the product of mass and velocity. Thus, the speed of the given particle will vary inversely with its mass. A proton with a greater mass will take a lower speed, while an electron with a lower mass will have a greater speed.

Q.2. Give the relation between de-Broglie wavelength and kinetic energy of an object.
Ans: The expression for the de-Broglie wavelength of an electron moving with kinetic energy \(\text {K}\),
\(\lambda = \frac{{\rm{h}}}{{\sqrt {2\,{\rm{mK}}} }}\)

Q.three. What was the de-Broglie hypothesis?
Ans: De-Broglie'south hypothesis states that all matter possesses both particle and wave-similar properties associated with it. He gave an equation that relates the wavelength of the given matter with its momentum.

Q.4. What is thermal de-Broglie wavelength?
Ans: The thermal de Broglie wavelength is equivalent to the boilerplate de Broglie wavelength of the gas particles in an ideal gas at the specified temperature.

Q.5. What are Matter waves?
Ans: According to De-Broglie, a wave associated with each moving particle is known equally a matter wave.