de-Broglie wave equation-Jayam chemistry learners

By V. Sai Kalyani

 de-Broglie equation-formula, derivation & significance:

If everyone is thinking alike, we cannot solve the hurdles of a problem. It seems true if we observe these hypotheses' hierarchy once. Italian physicist Francesco Maria Grimaldi discovered the wave phenomenon of light in 1665. But the uncertainty about light's nature was finally solved by Einstein's explanation of the photoelectric effect. Similarly, Neil Bohr succeeded in describing the structure of an atom with quantized electron orbits. But his stipulation of allowed stationary orbits was only a supposition until the discovery of the de-Broglie equation.

It is a mind map of the de-Broglie wavelength.

Table of contents:

1. What is de-Broglie equation?

2. De-Broglie wavelength formula

3. Derivation of de-Broglie equation

4. Experimental verification of de-Broglie equation

5. Significance of de-Broglie equation

6. Limitation of de-Broglie equation

7. Applications of de-Broglie equation

8. What is a matter wave?

9. MCQs on de-Broglie equation

10. Numerical problems on de-Broglie equation

 What is de-Broglie equation?

Louis de-Broglie, a French physicist, presumed that moving microscopic and macroscopic objects are waves. He introduced a word called 'matter wave' to describe the waves of material objects in motion. As a result, matter exhibits a dual character of both particle and wave. Moreover, he derived an empirical formula to measure the wavelength of matter waves in 1923 called the de-Broglie equation.

Consider an object having mass m will move with a velocity of v and has a matter wave with a wavelength of λ. Then by applying the de-Broglie equation, we have;

λ = h⁄mv

If the same object moves with the speed of light in a vacuum, then the wavelength of its matter wave can be;

λ = h⁄mc

Where, c = speed of light in the vacuum

(Or ) λ = h⁄m (3 x 108 m/sec)

The de-Broglie equation to enumerate the wavelength of matter waves is useful for microscopic particles such as electrons, protons, neutrons, atoms and molecules, positrons, etc., Due to their extreme low wavelength values, it is not applicable for macroscopic objects. Still, it proved the dual character of matter waves like light.

de-Broglie wavelength formula:

de-Broglie wavelength is the distance between two consecutive crests or troughs of a matter wave for a moving material object. It determines the distance covered by a moving material body having a specific mass and velocity.

Its SI unit is the meter, and the CGS unit is the centimeter.

De-Broglie wavelength is equal to the ratio of Planck's constant with a multiplicative product of the mass and velocity of a material object. 

λ = h⁄mv

Where,

λ = Wavelength of the matter wave

h = Planck's constant

m = mass of the object

v = velocity of the object

Momentum is a characteristic of a moving body that defines its quantity of motion as a multiplicative product of its mass and velocity. Hence, the de-Broglie wavelength varies inversely with the body's momentum.

de-Broglie wavelength ∝ 1⁄Momentum of body

(Or) 

λ = h⁄p

Where, 

p = momentum of body

As a result, heavier bodies with rational velocities own shorter de-Broglie wavelengths. Hence, the wave motion of a macroscopic object is much less significant. Conversely, microscopic particles possess longer de-Broglie wavelengths.

Derivation of de-Broglie equation:

De-Broglie presumed the wave nature of matter. And his empirical formula relied on Planck's quantum theory and Einstein's relation to define the wavelength of motion wave of matter.

The relationship between a photon’s energy and its wavelength following Planck quantum theory is as below;

E = hc⁄λ ------(1)

Where, E = Energy of light particle

According to Einstein’s mass energy relationship, we have;

E = mc2---------(2)

From eq (1) and (2), we have;

hc⁄λ = mc2

λ = h⁄mc

Where, c= Velocity of light in the vacuum

h = Planck’s constant

 m = mass of the particle

By substituting Planck's constant and speed of light values in the above equation, we get;

λ = 6.626 x 10-34joule sec⁄m x (3 x 108 m/sec)

λ = 2.208 x 10-42⁄m  meter

Hence, the de-Broglie wavelength varies inversely with the mass in case the material moves with the speed of light.

Experimental verification of de-Broglie equation:

Davisson and Germer, in 1927, proved the wave phenomenon of electrons which served as practical evidence for the de-Broglie equation. They used incandescent tungsten filament to produce a beam of electrons. And this electron beam was accelerated in an electric field. Then the electron ray was allowed to fall on a nickel crystal surface to split in different directions. It is known as grating. As a result, rounded dark and bright diffraction rings of electron beams formed on a photographic plate. These concentric electron beam rings resembled the X-ray diffraction pattern. It confirmed the wave motion of the electron.

It is an experimental verification of the dual nature of the electron.

If V is the potential difference applied in the electric field to accelerate an electron of charge e. Then the kinetic energy acquired by the electron is

KE of the electron = Ve

1⁄2 mv2 = Ve

v = (2Ve⁄m)1/2

From de-Broglie equation, we have;

λ = h⁄mv

By substituting the value of 'v' in the avove equation, we get;

λ = h⁄m(2Ve⁄m)1/2

On solving, we will get;

λ = h⁄(2Vme)1/2

By substituting the values of Planck’s constant, charge, and mass of the electron in the above equation, we get;

λ = 6.626 x 10-34 joule sec⁄{2(1.602 x 10-19 coulombs)(9.1 x 10-31Kg x V volt}1/2

λ = 12.265 x 10-10 metre ⁄(v)1/2

The above equation uses the potential difference applied to accelerate the electron in the electric field to determine its de-Broglie wavelength.

For example- If the potential difference of accelerating electrons in the electric field is 54 volts, then by using above equation its de-Broglie wavelength can be;

λ = 12.265 x 10-10 metre ⁄(54)1/2

λ = 1.66 A0

Hence, an electron accelerating with 54 volts applied potential difference in the electric circuit shows a de-Broglie wavelength of 1.66 angstroms. This wavelength comes under X-ray region. Hence, the diffraction grating rings of the electron and the X-rays resemble one another.

Significance of de-Broglie equation:

1. It successfully measured the wavelengths of moving microscopic particles. And it introduced the dual character of matter.

For example: If an electron moves with 100 m/sec, then the wavelength of the mobile electron wave is below;

λ = 6.626 x 10-34 joule second⁄(9.1 x 10-31 Kg) x 100 m/sec

λ = 0.728 x 10 -5 meter

2. It described Neil Bohr’s quantized angular momentum condition mathematically. 

According to Bohr’s atomic model, an atom can have an infinite number of stationary orbits. But the electron rotates in permitted stationary orbits where the electron’s angular momentum is an integral multiple of h/2π. Hence, all stationary orbits around the nucleus of an atom are not suitable for accommodating the electrons.

Consider an electron with mass m and velocity v rotates in a circular orbit having radius r. Then the angular momentum of the electron = mvr

Bohr’s quantized angular momentum condition is;

mvr = nh⁄2π

Where, n is an integer

On rearranging the above equation, we get;

2πr = nh⁄mv

Applying de-Broglie equation, we get;

2πr = nλ

The above equation shows that an allowed stationary orbit with a circumference of 2πr holds an integral number of de-Broglie wavelengths of an electron wave. Then the electron wave is said to be in phase. For an in-phase electron wave, the two ends of regular crests and troughs series meet to give a closed circular arrangement resembling the electron’s orbit. It prevents the loss of electrons’ energy during rotation.

When the stationary shell with 2πr circumference holds a fractional number of de-Broglie wavelengths of an electron wave, then it is said to be out of phase. Due to the irregular pattern of crests and troughs in the electron wave, the motion of the electron is not circular. As a result, the electron loses energy during rotation. Hence, these stationary orbits are not allowed to accommodate the electron.

Limitation of de-Broglie equation:

De-Broglie equation is much less applicable to macroscopic objects which we see in our daily life. It is due to the negligible wavelengths of their matter waves. Hence, we do not observe massive bodies propagate like a wave.

For example- a football of weight 0.01 Kg moves with a velocity of 100 m/s. The wavelength of the ball’s motion wave from the de-Broglie equation is as below;

λ = 6.626 x 10-34 joule second⁄(0.01) kg x 100 m/sec

λ = 6.626 x 10-34 meter

This wavelength is much smaller than any light radiation of the entire electromagnetic spectrum. Hence, any practical technique cannot measure it. So, the de-Broglie equation is invalid for visible objects with considerable mass.

Application of de-Broglie equation:

  1. An Electron microscope developed based on the wave nature of electrons help in the study of microscopic particles at atomic and molecular levels with greater magnification and finer details. Therefore, an electron microscope is of great use in scientific research.
  2. The wave nature of electrons helps to research the surface structure of solids.

What is a matter wave?

Matter waves are waves that associate with material particles. All material objects exhibit wave character while moving, irrespective of size and mass. As a result, both the microscopic and macroscopic bodies are waves in motion. And the motion waves are called matter waves.

Matter waves have no associated electric and magnetic fields. And they are not emitted from a source. They only associate with the moving material's particles.

Matter waves cannot pass through the vacuum. They require a medium for their propagation. Hence, their velocity is less than the speed of light. And they exhibit different velocities based on the density of the propagating medium.

Matter waves of larger objects have diminutive wavelengths. Hence, they have no prominence in our routine life. 

MCQs on de-Broglie equation:

1. Which of the following microscopic particle has the longest de-Broglie wavelength? Provided all of them move with the same velocity.

  1. Ozone

  2. Proton

  3. Electron

  4. Neutron

Answer:
Electron
Explanation:
Electron is the lightest particle than the remaining ones. Hence, it has a longer de-Broglie wavelength when all the above moves with the same velocity.


2. De-Broglie equation did not apply to_______________

  1. Positron

  2. Proton

  3. Photon

  4. Molecules

Answer:
Photon
Explanation:
Photon is a light energy particle comprised of electromagnetic waves while moving. The de-Broglie equation applies to matter waves but not electromagnetic waves. Consequently, the de-Broglie equation is inapplicable for photons.


3. Who introduced matter waves?

  1. Heinrich Hertz

  2. James Clerk Maxwell

  3. Max Planck

  4. de-Broglie

Answer:
de-Broglie
Explanation:
Louis de-Broglie, a French physicist, invented the phrase matter waves. Matter waves associate with mobile material particles.


4. If the velocity of a ball suddenly rises to 25 times, what will be its wavelength?

  1. Increases

  2. Decreases

  3. no change

  4. None of the above

Answer:
decreases
Explanation:
De-Broglie equation interprets the inversely proportional relationship between the wavelength and velocity of an object. Hence, the wavelength of a matter wave decreases with the rise in the speed of the moving body.


5. What is the relationship between the de-Broglie wavelength and the momentum of a material object?

  1. Both of them vary directly

  2. Both of them vary inversely

  3. Momentum is constant at all wavelengths

  4. equal

Answer:
Both of them vary inversely
Explanation:
De-Broglie wavelength and momentum of an object are inversely related quantities. As a result, larger objects with appreciable momentum possess infinitesimal small wavelengths.


6. What is the requirement of an in-phase electron wave?

  1. mvr = nh⁄2π

  2. nhr = mv

  3. 2πr= nh

  4. 2πr = nλ

Answer:
2πr = nλ
Explanation:
An electron wave is said to be the in- phase when the circumference of the stationary orbit is an integral multiple of wavelengths of the electron wave. It gives a closed spherical electron wave pattern in which the edges of a wave meet to prevent the loss of an electron's energy by rotation.


7. What is the CGS unit of de-Broglie wavelength?

  1. nm

  2. pm

  3. cm

  4. μm

Answer:
cm
Explanation:
De-Broglie wavelength gives the distance covered by a moving material body having a definite mass and velocity. As a consequence, the centimeter is its CGS unit.


8. Name the experiment that stood as practical evidence of the de-Broglie hypothesis?

  1. Davison and Germer electron diffraction experiment

  2. Einstein's quantum mechanical approach

  3. Electron scintillation experiment

  4. Optical diffraction method

Answer:
Davison and Germer electron diffraction experiment
Explanation:
De-Broglie's hypothesis relies on the wave phenomenon of movable material objects. But it did not have any experimental verification until Davisson and Germer's electron beam diffraction method proved it in 1927.


9. de-Broglie equation applies to _________________

  1. only microscopic particles

  2. only macroscopic particles

  3. Both microscopic and macroscopic particles

  4. Electromagnetic radiations

Answer:
Both microscopic and macroscopic particles
Explanation:
All material substances, like microscopic and visible, show wave motion when moving following the de-Broglie equation.


10. When does an object behave as a wave following the de-Broglie hypothesis?

  1. At rest

  2. In an electric field

  3. In magnetic field

  4. In motion

Answer:
In motion
Explanation:
The wave phenomenon of motile substances is the keynote of de Broglie's proposal. So, all moving objects exhibit wave nature.


11. If momentum of an iron wheel is 250 Kg m/s. Then what will be its de-Broglie wavelength?

  1. 0.023 x 10-34

  2. 2.605 x 10-36

  3. 7.9 x 10-23

  4. 0.026 x 10-31

Answer:
2.605 x 10-36
Explanation:

λ= h⁄p
λ = 6.626 x 10-34⁄250
λ = 2.605 x 10-36 meter


12. What is the nature of a proton by de-Broglie theory?

  1. a sub-atomic particle

  2. An electromagnetic wave

  3. Both particle and wave

  4. Microscopic particle

Answer:
Both particle and wave
Explanation:
Proton is a sub-atomic particle located at the atom's center that exhibits particle and wave characteristics by de Broglie's hypothesis.


13. de-Broglie hypothesis proves_______________

  1. Dual nature of photon

  2. Dual nature of matter

  3. Dual nature of electron

  4. None of the above

Answer:
Dual nature of matter
Explanation:
Dual nature means particle and wave nature of substances (or) of the matter is the motto of the de-Broglie proposal.


14. Which of the following microscopic particle possesses matter waves?

  1. Electron

  2. Atom

  3. Molecule

  4. All

Answer:
All
Explanation:
Invisible microscopic matter such as electrons, protons, neutrons, atoms, and molecules move as a wave following the de-Broglie theory.


15. What is the velocity of matter waves?

  1. 3 x 108m/sec

  2. Less than 3 x 108m/sec

  3. Greater than 3 x 108m/sec

  4. None of the above

Answer:
Less than 3 x 108m/sec
Explanation:
Matter waves cannot propagate in space as they comprise material particles. Consequently, their velocity is less than the speed of light in a vacuum.


16. The electron diffraction rings of the electron beam match closely with__________

  1. Uv light

  2. Visible light

  3. X-rays

  4. Radio waves

Answer:
X-rays
Explanation:
Electron diffraction experiments produce intense diffraction patterns when the applied potential difference in the electric field remains at 54 volts. It gives matter waves with a wavelength of 1.65 Angstrom. Hence, the electron diffraction rings match closely with the X-ray rings as their wavelength ranges are similar.


17. Which of the following equipment invention is the application of the de-Broglie hypothesis?

  1. Electron microscope

  2. Compound microscope

  3. Spectroscope

  4. Telescope

Answer:
Electron microscope
Explanation:
Electron microscopes rely on the wave phenomenon of the electron. And it is an application of the de-Broglie hypothesis.


18. The basis of electron microscope is_______________

  1. The electron is a negatively charged microscopic particle

  2. Velocity of electron

  3. Wave nature of electron

  4. Participation of an electron in chemical reactions

Answer:
Wave nature of electron
Explanation:
An electron accelerated to a potential difference of v volts to get finer details of microscopic particles using the electron microscope. Hence, the electron's wave nature is employed to magnify invisible miniature particles of the universe for scientific research.


19. Which of the following phenomenon confirms the wave nature of electrons?

  1. Deflection

  2. Diffraction

  3. Reflection

  4. Interference

Answer:
Diffraction
Explanation:
Diffraction is a wave phenomenon showing the spreading of waves when passes through an aperture.


20. How does the wavelength of an electron wave relates to potential difference of electric field?

  1. λ = V x m

  2. V = (λ⁄2im)1/2

  3. λ = h⁄(2Vme)1/2

  4. V = λ⁄2hm

Answer:

  • λ = h⁄(2Vme)1/2


    • 21. Matter waves associates with_________________

    1. moving neutral particles

    2. charged particles in motion

    3. moving material particles

    4. matter at rest

    Answer:
    Moving material particles
    Explanation:
    Matter waves are particle waves that associate with motile objects.


    22. Matter waves are_________________

    1. Transverse waves

    2. Electromagnetic waves

    3. Mechanical waves

    4. None of the above

    Answer:
    None of the above
    Explanation
    Matter waves are waves that associate with moving material particles. Hence, they are neither mechanical nor transverse and electromagnetic.


    23. _____________ is useful for grating in Davisson and Germer electron diffraction experiment

    1. Nickel crystal

    2. Gold plate

    3. Alpha particle

    4. Tungsten filament

    Answer:
    Nickel crystal
    Explanation
    Nickel crystal scatters the accelerated electron beam in various directions, known as grating.


    24. Particle nature of electron confirmed by-------------------------experiment

    1. Electron diffraction experiment

    2. Scintillation experiment

    3. Photoelectric cell

    4. Gold foil experiment

    Answer:
    Scintillation experiment
    Explanation
    When the electron strikes the Zinc sulfide-coated plate, it produces a spot of light called scintillation. This experiment proves striking electron act as a particle.


    25. Which of the following proposal of Bohr's atomic model explained by the de-Broglie equation?

    1. Energy quantization

    2. Fine structure splitting

    3. Quantization of angular momentum

    4. Atomic spectrum

    Answer:
    Quantization of angular momentum
    Explanation
    Bohr's atomic model states that the electrons incur only those stationary orbits in which the circumference of the electronic shell is a whole multiple of h/2π. And de-Broglie equation mathematically proves the quantization of the electron's angular momentum in Bohr's orbit.

    Numerical problems on de-Broglie equation:

    Problem-1: 

    What will be the wavelength of a ball of mass 0.01kg moving with a velocity of 50 m/s?

    Solution:

    Mass of ball = 0.01 kg

    Velocity of moving ball = 50 m/s

    The formula to calculate the wavelength of moving ball is;

    λ = h⁄mv

    λ = (6.626 x 10-34)⁄0.01 x 50

    λ = 13.252 x 10-34meter

    Problem-2:

    Find the mass of a yellow-colored emission of sodium atom during the flame test when the wavelength of yellow light is 580 nm.

    Solution:

    Wavelength of emitted yellow light from sodium atom = 580 nm

    Velocity of light = 3 x 108 m/sec

    Formula to calculate the mass of yellow emission of sodium atom is;

    m = h⁄λv

    m = (6.626 x 10-34)⁄(580 x 10-9meter) (3 x 108m/sec)

    m = 0.003808 x 10-33m

    m = 3.808 x 10-36kg

    Problem-3:

    Calculate the wavelength of the electron if its kinetic energy is 10-25 joule.

    Solution:

    Kinetic energy of electron = 10-25 joule

    1⁄2mv2 = 10-25

    v = {(2 x 10-25joule) ⁄ (9.1 x 10-31kg)}1/2

    v = 468.8 m/sec

    The formula to calculate the wavelength of electron is;

    λ = h⁄mv

    λ = (6.626 x 10-34js)⁄(9.1 x 10-31kg) (468.8 m/sec)

    λ = 155 x 10-8meter

    Problem-4:

    What is the frequency of an unknown particle wave if the particle’s kinetic energy is 5 x 10-24 joule?

    Solution:

    Kinetic energy of particle = 5 x 10-24 joule

    1⁄2mv12 = 5 x 10-24joule

    mv12 = 10 x 10-24joule

    mv12 = 10-23joule

    According to Planck quantum theory, we have;

    E = hν1 = hc⁄λ

    It implies,

    Wavelength (λ) = Velocity of particle (v1)⁄Frequency (v)

    According de-Broglie equation, the formula to calculate the wavelength of particle wave is;

    λ = h⁄mv1

    By comparing the above two equations, we get;

    v1⁄v = h⁄mv1

    v = mv12⁄h

    v = 10 -23 joule⁄6.626 x 10-34 js

    v = 0.15 x 1011Hz

    Problem-5:

    Calculate the de-Broglie wavelength of the helium atom that moves with 800 m/s.

    Solution:

    Velocity of helium atom= 800 m/s

    Mass of a single helium atom is;

    mHe = Gram atomic mass of Helium⁄Avogadro's number

    mHe = 4⁄6.023 x 1023

    mHe = 0.664 x 10-23kg

    The formula to calculate the de-Broglie wavelength of helium atom is;

    λ = h⁄mv

    λ = 6.626 x 10-34joule second⁄(0.664 x 10-23kg) x (800 m/s)

    λ = 0.0124 x 10-11 meter

    Problem-6:

    Calculate the kinetic energy of a moving sub-atomic particle with mass 9.1 x 10-31 kg, and the wavelength of its matter wave is 4.8 A0?

    Solution:

    Mass of the sub-atomic particle = 9.1 x 10-31 kg

    Wavelength = 4.8 A0

    De-Broglie wavelength of sub-atomic particle matter wave is;

    λ = h⁄mv

    v = h⁄mλ

    v = 6.626 x 10-34 joule second⁄(9.1 x 10-31 kg) (4.8 x 10-10 meter)

    v = 0.151 x 107m/s

    Kinetic energy of moving sub-atomic particle is;

    KE = 1⁄2mv2

    KE = 1⁄2(9.1 x 10-31kg) (0.151 x 107m/s)2

    KE = 0.103 x 10-17joule

    Problem-7:

    An electron beam can undergo diffraction by crystals. Through what potential should a beam of electron be accelerated so that the wavelength becomes equal to 1.54 A0?

    Solution:

    The wavelength of accelerated electron beam = 1.54 A0

    According to de-Broglie equation, we have;

    λ = 12.265 x 10-10 meter⁄(V)1/2

    (V)1/2 = 12.265 x 10-10⁄1.54 x 10-10

    V = (7.96)2 volts = 63.36 volts