As per the de Broglie’s hypothesis, it can be said that, a particle is considered as a group, or packet of waves. The constituent waves are of nearly equal wavelengths, whose values are centered around the de Broglie wavelength 1.. = h/p. The wave packet is formed due to the

superposition of waves. The amplitude of wave packet is zero everywhere in space, except only over a small region in which the particle is located. Essentially, the wave packet’s extent is limited over this small region of space, but travels with a velocity equal to its group velocity, which is equal to the particle velocity.

Because that the wave packet is spread in space, it implies that the position of the particle cannot be pin pointed or specified accurately. It can only be said that, the particle is somewhere within the region over which the wave packet is spread.

Let the motion of the particle be along X-axis. If the wave packet extends over a finite distance Ax, along the axis of motion, then the two points at which the amplitude of the wave group becomes zero successively, will be separated by a distance Ax (Figure 9.6).

**FIGURE 9.6** **Spatial extension of wave packet**

Two points where the amplitude become S zero are the nodes. Any measurement made to find the position of the particle will have a minimum error equal to the distance between the two nodes. This is inevitable due to the wave nature of matter.

The amplitude of the wave packet is given by

2A cos [ ( dw /2 )t – (dk /2 ) x ]

At nodes, the amplitude is zero.

2A cos [ dw / 2 t –dk /2 x ] = 0

2A = 0 , Hence cos [ dw /2 t –dk /2 x ]=0

Or dw / 2 t – dk /2 x = π /2 , 3 π /2 ,5 π/ 2 ….. (2 n +1) π /2

If x_{1} and x_{2} be the two positions at which the nodes are formed, then, for the first node

(Dw /2 ) t –dk /2 x_{1} =(2n+1) π/2

Since, there is a change of a phase 1t, between the successive nodes, the equation for second node is

(Dw /2 ) t –dk /2 x_{2} =(2n+1) π/2 +π

Subtracting equation (1) from equation (2), we have

Dk/2 (x_{1} –x_{2}) =π

X_{1}-x_{2} = dx

Dk/2 (dx) = π

Dx = 2π /dk => dx = 2π /dk

which is the fundamental error in the measurement of the position of particle.

But, we know that k = 2π /λ = 21tp _{x /} h

(by using de Broglie equation A.= h /p _{x}),

where, P _{x} is the momentum along x-axis and his the Planck’s constant

or ∆k =2π∆p _{x} / h

where ∆p _{x} represents the fundamental error in the measurement of the momentum of the particle. ·

Substituting for ∆k in equation (3) from equation (4), we have

∆x =h /∆ p _{x}

Or ∆x∆ p _{x} =h

But as per Fourier analysis, the width 6.x of a single wave group which represents the

superposition of waves, whose propagation constants vary in the range &, obey the relation

∆x =1 / ∆k

Whereas, equation (3) is obeyed, when the superposition is due to only two wave

trains. Hence for general superposition, equation (6) is valid.

. . Substituting for ∆k from equation (6) in equation (4), we have

1 / ∆x =2π (∆ p _{x}) /h

∆x ∆p _{x }= h / 2π

Since, the fundamental errors involved in the measurement of x and P x could as best be equal to, or greater than, but will never be less than ∆x and ∆P x respectively, the equation is more rationally expressed as

**9.5.1 Statement of Heisenberg’s Uncertainty Principle**

In any simultaneous determination of the position and momentum of the particle, the product of the corresponding uncertainties inherently present in the measurement is equal to, or greater than (h/2π).

The Heisenberg uncertainty principle could also be expressed in terms of the uncertainties involved in the measurement of physical variables-pair, such as energy and time,

angular displacement (6) and angular momentum (L). Both the pairs are conjugate in the sense that, their respective products have the dimensions of Planck’s constant

Thus, we have the uncertainty relations

∆E ∆t ≥ h /2π

And ∆ L ∆ θ ≥ h /2π

where, the notation l1 is association with the respective variables, indicates the minimum uncertainty involved in the measurement of the corresponding variables.

**9.5.2 Physical Significance of Heisenberg’s Uncertainty Principle**

In classical mechanics, by the very definition itself, a particle occupies a definite place in space, and possesses a definite momentum, knowing the particle’s position and momentum at any given instant of time, it is possible to evaluate its position and momentum at any later point of time, and the trajectory of the particle could be continuously traced. but, such a notion breaks down in atomic scales, wherein a new mechanics, known as the wave mechanics, or quantum mechanics takes over, for which one of the fundamental principle is the Heisenberg’s uncertainty principle.

As we realise from the principle, it is impossible to determine simultaneously both the position and momentum of a particle accurately. Any efforts made to make the measurement of the position of the particle such as an electron very accurately, results in a large uncertainty in the measurement of momentum and vice versa. In effect, irrespective of the level of sophistification arranged for the measurements, the measured values cannot have accuracies exceeding what is permitted under the condition

∆x ∆p x ≥ (h /2π)

where, ∆_{x} and ∆p = are the uncertainties in the measured values of the position and momentum of the particle.

The consequences arising from the above conditions, i.e., the uncertainty principle are as follows :

An increase in accuracy in the measurement of position means a decrease in the value of ∆x.

As per the above condition, a decrement in ∆lx must simultaneously result .in a proportionate increment in the value of ∆p _{x}, since the right hand side of the identity is a constant. The physical significance of the above arguments is that, one should not think of the exact position, or, an accurate value of momentum of a particle. Instead, one should think of the probability of finding the particle at a certain position, or of the probable value for the momentum of the particle. The estimation of such probabilities are made by means of certain mathematical functions, named probability density functions in quantum mechanics.

**9.5.3 Impossibility of Existence of Electrons in the Atomic Nucleus**

We know the theory of relativity· that, the energy E is expressed as

Also, we know that, momentum p is given by

P^{2} = m_{0}^{2}v^{2} / (1- v^{2} /c^{2} ) = m_{0}v^{2} c^{2} / (c^{2} –v^{2} )

Multiplying by c^{2}, we have

P^{2} c^{2} = m_{0}v^{2} c^{2} / (c^{2} –v^{2} )

Subtracting equation (2) by equation (1), we have

E^{2} =p^{2} c^{2} = m_{0}v^{2} c^{2} / (c^{2} –v^{2}) / (c^{2} –v^{2})

E^{2 }=p^{2} c^{2} = m_{0}v^{2} c^{4}

E=p^{2} c^{2} = m_{0}v^{2} c^{4}

^{ }

Heisenberg’s u11certainfy principle states that

∆p x ∆x ≥ h / 2π

We know the diameter of the nucleus is of the order of 10^{-14} m. If an electron is expected to exist inside the nucleus, then the uncertainty in its position ∆.x must not exceed the size of the nucleus

i.e; ∆x ≤10 ^{-14} m

In such an event, the uncertainty in its momentum as per equation (4) becomes

∆p x ≥ h / 2π ∆x

∆p x ≥ 6.63 x 10 ^{-34} / 2π x 10 ^{-14}

∆p x ≥1.055 x 10 ^{-20} Ns

which is the uncertainty in the momentum of the electron, in which case, the momentum of the electron must at least be equal to the uncertainty in the momentum, i .e., 1.055 x 10^{– 20}

p x ≥ 1.055 x 10 ^{-20} Ns

Now, from equation (3)

E^{2 }=p^{2} c^{2} = m_{0}v^{2} c^{4} (P^{2} + m_{0}^{2}v^{2)}

we know, the rest mass of the electron

m_{0} = 9.11 x 10^{-31} kg

Now, by making use of the condition in equation (6), we can say that in order that the

electron may exist within the nucleus, its energy E must be such ·that

E ^{2} ≥ (3 x 10 ^{8})^{2} [(1.055 x 10^{-20})^{2} + (9.11 x 10)^{2} (3 x 10^{8})^{2} ]

i.e., E ^{2} (3 X 10^{8})^{2}[1.113 X 10^{-40} + 7.4692 x 10^{-44} )

Since the second term in the bracket is smaller by more than 3 orders of magnitude compared to the first term, it is neglected.

Taking the square roots on both sides, and .simplifying, we get the condition

E ≥ 3.1649 x to^{-12} J

Expressing in eV, we get

E ≥ 20 MeV

This means to say that, in order that an electron may exist inside the nucleus, its energy must be greater than or equal to 20 MeV. But, the experimental investigations on β-decay reveal that, the kinetic energy of the β-particles (which are same as electrons) is of the order of 3 and 4 MeV.

This clearly indicates that, electrons cannot exist within the nucleus.