2.10: Quantum Mechanical Model of the atom:

1. de. Broglie’s equation and Heisenberg’s Uncertainty Principle established that electron has a dual nature. It can be regarded both as a particle and as a wave. Therefore, it is not logical to assume that electrons move in well defined orbits. This led to the introduction of the concept of probability in describing the atomic systems and gave birth to a new model known as Wave mechanical model of an atom or Quantum mechanical model of atom.

2.10.1: l. de. Broglie’s equation:

In 1924 by the French physicist Louis de Broglie suggested that like photons, all material particles such as electron, proton, neutron, atom, molecule, a stone or a ball (i.e., microscopic as well as macroscopic objects) in motion have a dual nature, i.e., both exhibit wave character as well as particle character. The wave associated with a particle is called a matter wave. The wavelength associated with a moving particle is given by the following relation which is called de- Broglie’s equation;  λ =h/mv = h/p

Where, λ= wave length of moving particle, m = mass of moving particle and h= Planck’s constant.

All the properties of the electron can be explained by its wave –particle duality.

From de. Broglie’s equation λ α 1/mv [ as h is constant]

i.e,  wavelength α       1                              .

momentum

Thus, the wavelength of the wave associated with a moving particle e.g. electron is inversely proportional to its momentum.

Fig. 2. 38 de-Broglie’s stationary wave quantised orbits

2.10.2: Derivation of de Broglie’s equation:

According to Max Planck’s quantum theory ,  ……………(1)

Where E = energy,

h = Plank’s constant (6.62607 x 10^-34 J s),

ν= frequency

Einstein’s famous equation relating  to the special theory of relativity, ……………..(2)

Where, E = energy,

m = mass,

c = speed of light

Equate these two equations and recall the expression relating the frequency and wavelength of a photon, we get

By analogy, de Broglie argued that a particle with non-zero rest mass m and velocity v would have a wavelength given by

Since mv=p, where p is the particle’s momentum, the now famous de Broglie relation becomes

Equation (6) and (7) is known as de-Broglie’s equation.

2.10.3: Experimental verification:  Davisson and Germer in 1927, accelerated the speed of a stream of electrons emitted from a tungsten filament by applying a strong electric field. When these electrons were made to strike the nickel crystal (worked as grating), concentric dark and bright rings were formed on the screen. This diffraction pattern was similar to the pattern observed with X-rays. Since, X-rays are electromagnetic waves (wave character), therefore, electrons must also have the wave

nature.

Fig.2.38 Electron diffraction experiment by Davisson and Germer.  

2.10.4: Significance of de. Broglie’s equation: On the basis of de Broglie equation, the following

conclusions can be drawn:

(i) Lighter the particle, greater is its de Broglie wavelength for certain velocity.

(ii) Faster the particle moves, smaller is its de Broglie wavelength.

(iii) The de Broglie wavelength is independent of the nature of the particle.

(iv) The de Broglie equation is applicable to microscopic particles such as electron, proton, atom, molecule, etc., and has no significance for heavier particles. Let us apply the equation to a moving electron and a moving ball weighing 200 g or 0.2 kg respectively,  both travelling with the same speed, 2500c ms^-1.

For moving electron:

λ = h /mv =    6.625 x 10 ^-34  (J-S)                      =  0.000291 x 10 ^-3   m

9.1 x 10 ^-31 (Kg ) x 2500 (ms^-1 )

For moving ball:

λ = h /mv =      6.625 x 10 ^-34  (J-S)        = 0.01325 x 10 ^-34     m

0.2 (kg) x 2500 (ms^-1 )

Thus, the wavelength is too small in the case of  moving ball (heavier particles ) which are difficult to measure and has no significance.

 

2.10.5  Derivation of Bohr’s postulates of Quantisation of Angular Momentum from de-Broglie’s relation:

Bohr’s condition, that the angular momentum is an integer multiple of h/2π, was later reinterpreted in 1924 by de Broglie as a standing wave condition. Electrons can exist only in locations where they interfere constructively.

Allowed orbits are those in which an electron constructively interferes with itself. Not all orbits produce constructive interference and thus only certain orbits are allowed (i.e., the orbits are quantized). By assuming that the electron is described by a wave and a whole number of wavelengths must fit along the circumference of the electron’s orbit, we have the equation:

nλ=2πr…………. (1)

or, λ=2πr/ n…………. (2)

Where, r = radius of the orbit and  n = an integer

According to de-Broglie’s equation λ=h/mv

 

Where, m = mass of electron, v = velocity of the electron

Thus, 2πr/ n = h/mv ………….. (3)

or,  mvr =nh/2 π …………(4)

Relation  (4) is same as the relation obtained from  Bohr’s theory for Quantization of angular Momentum

as the condition for allowed orbits:

mvr=nh/ 2π,     Where, n=1,2,3….

 

Fig. 2.39 Constructive interference and destructive interference

2.10.7:  Derivation of K.E of a moving electron and de-Broglie’s wave length associated with it:

According to de-Broglie’s relation

λ = h/p ……………..(1)

or,  λ = h/mv ………………(2)

Where λ = wave length,

h = Plank’s constant (6.62607 x 10^-34 J s),

p = momentum of the moving electron

m = mass of the moving electron,

v = velocity of the  moving electron.

Kinetic energy of the moving electron is given by

From E=1/2 m v² ………………… (3)
or, 2E = m v²
Multiplying both side by m, we get

2Em = (m v) ² 
or, 2Em = p²

or, p = √(2Em)  ……………(4)

from equation (1) and (4) we get
h/ λ  = √ (2Em)

or, λ =       h   .   …………………(5)

(2Em)

Relation (5) gives the relation between of K.E of a moving electron and de-Broglie’s wave length associated with it.

Table 2.9 Distinction between wave and particle:

Sl. no.	Wave	Particle
01.	A wave does not have a localised position. However, it is spread out in space, i.e., it propagates in space.	A particle occupies a well defined position in space, i.e .it has always a localised position.
02.	Two or more waves can exist simultaneously in a given  region of space.	Two or more particles cannot exist simultaneously at a given position in space.
03.	The two waves present in the same region may undergo constructive or destructive interference, i.e., they may superimpose to form a resultant wave which can be  smaller or larger than individual waves	Particles do not show the phenomenon of interference.
04.	The wavelengths of these waves are given by the relation, E= h γ, i.e., Planck’s equation.	The wavelengths of these waves are given by the relation, λ = h/mv, i.e de Broglie’ s equation.

 

 

Table 2.10. Distinction between Electromagnetic waves and Matter waves:

Sl. no.	Electromagnetic waves	Matter waves
01.	These waves are emitted from a source.	These waves are always  associated with moving particles. These are not emitted from any source.
02.	All electromagnetic waves travel with the same velocity, i.e., velocity of light.	These waves travel with different velocities and their velocities are generally less than the velocity of light.
03.	The waves are associated with oscillating  electric and magnetic fields, perpendicular to each other and to the direction of propagation of radiation.	These waves are not associated with  electric and magnetic vectors.
04.	The waves do not require any material medium for propagation. These can travel in vacuum.	These waves require a material medium for propagation. These waves cannot travel through vacuum.
05.	The wavelengths of these waves are given by the relation,  λ = c/ν where, v is the frequency and  c is the velocity. The wavelengths are much larger	The wavelengths of these waves are given by the relation,  λ = h/mv where, m is the mass of the particle moving with velocity v, h is Planck’s constant. The wavelengths are much shorter.
06.	The energies are quantized	The   energies are    not quantized.
07.	These waves can be absorbed, evolved   and propagated.	These waves do not radiate. These are neither absorbed nor evolved.

 

 

 

 

Numerical Problems:

 

01: What is the wavelength of an electron (mass = 9.11 x 10¯³¹ kg) traveling at a speed of 5.31 x 10⁶ m/s?

(Ans. 1.37 x 10^¯10 m)

02: Calculate the wavelength (in nanometers) of a H atom (mass = 1.674 x 10^-27 kg) moving at 698 cm/s.

(Ans. 150. Nm )

03: Calculate the wavelength of an object weighing 100.0 kg and moving at 160 km/hour. [1.49 x 10^-37 m ]

04: An atom of helium has a de Broglie wavelength of 4.30 x 10¯¹² meter. What is its velocity? (2.32 x 10⁴ m/s)

05: Calculate the velocity of an electron (mass = 9.10939 x 10¯³¹ kg) having a de Broglie wavelength of 269.7

pm (2.697 x 10⁶ m/s )

06: What is the de Broglie wavelength (in nm) of a molecule of buckminsterfullerene (C₆₀), moving at a speed of

100.0 m/s? (5.537 x 10¯³ nm )

07: Calculate (a) the de Broglie wavelength of an electron moving with a velocity of 5.0x 10⁵ ms^–1 and (b)

relative  de  Broglie wavelength of an atom of hydrogen and atom of oxygen moving with the same velocity .

(h = 6.63 x 10^–34 kg m² s^–1)

08: Calculate the accelerating potential that must be applied to a proton beam to give it an effective wavelength

of  0.005 nm.                                                                               [32.85 volt ]

9. A bacterium moving across a Petri dish at 3.5 µm/s has a de Broglie wavelength of 1.9 x 10 ^-13 m. What is

the  bacterium’s mass?

10. A baseball traveling horizontally at 32.0 m/s has a de Broglie wavelength of 1.06 x 10 ^-34 m.
11. What is the mass of the baseball?
12. Why does the baseball show very little wave behavior characteristics?
13. If an electron and a proton travel at the same velocity, which has the shorter wavelength? me =9.11 x10 ^-31

kg  and mp =1.6725 x10 ^-27 kg.

12. A proton (m = 1.67 x 10 ^-27 kg) with a de Broglie wavelength of 4.00 x 10 ^-14 m is moving at an unknown

velocity.  a) What is the proton’s velocity?

1. What is the proton’s momentum?
2. Why does the proton show very little particle behavior characteristics?
3. A moving particle is associated with wavelength 5 x10 ^-8m. Calculate its new wavelength when its

momentum is reduced to half of its value.  [10 ^-7 m ]