Heisenberg

HEISENBERG’S UNCERTAINITY PRINCIPLE Introduction When we are studying a large moving object say a planet, then we can follow its definite path on which it travels. If we know its initial position and momentum, then we can predict its position and momentum at any other time. But this is not possible for electron, proton and neutron which are microscopic particles. Heisenberg has given a principle in this connection. He says that it is impossible to measure simultaneously both the position and momentum of a microscopic particle with accuracy or certainity. Mathematically this principle can be put as follows: ∆X x ∆P ≥

𝒉 𝟒𝝅

∆X = Uncertainity in the position ∆P = Uncertainity in the momentum These two uncertainities are inversely proportion to each other. So, if position of the microscopic particle is known with more accuracy, then there will be more uncertainity in its momentum and vice versa. Physical concept of uncertainity principle: In order to know the position of an object, we throw the photons of light upon them. If we want to have the idea for the position of electron, then the photons of x-ray region have to be used because their wavelength are very small and the possibility for the hitting of electron is there. During the hitting, the photons transfers some of its energy to the electron. Therefore, the velocity and hence the momentum of electron changes. If we use the photons of longer wavelength say of visible region, the velocity and the momentum will not change appreciably because longer wavelengths rarely find chance to hit the electron. But its position cannot be determined because object will not be visible. Keep it in mind that, the uncertainity is not due to lack of better techniques for the measurements of position and momentum. It is due to the reason that we cannot observe the microscopic objects without disbursing them. Uncertainity principle is not applicable to stationary state the velocity of an electron is zero. As a result, position of an electron can be determined accurately. Mathematical form of uncertainity principle: We have to consider an hypothetical experiment in which we can measure the position and velocity of an electron. Following diagram shows an arrangement in this respect.

A photon from a source of 𝛾-rays or x-rays with energy ‘ℎ𝜐’ strikes the electron at the point P. When the electron scatters this photon into the microscope in a direction making an angle 𝛽 with the xaxis, the electron will receive some momentum, from the photon along x-axis. Scattered photon can enter the microscope anywhere within the angle ‘2𝛼′. Its contribution to x-component of the momentum of electron is, 𝟐𝒉 𝒔𝒊𝒏𝜶 𝝀

∆Px = 2Psin𝜶 = Because 𝝀

=

……(1)

𝐡 𝐦𝐯

Rayleigh’s equation for the resolving power can be used to find the accuracy with which an object can be located by a microscope and is given by, 𝝀

∆X = 𝟐𝒔𝒊𝒏𝜶

……(2)

∆X = Distance between two points which can just be resolved by the microscope.

𝝀 = Wavelength of photon. Multiplying equation (1) with (2) 𝝀

∆X . ∆Px = 𝟐𝒔𝒊𝒏𝜶 .

𝟐𝒉 𝒔𝒊𝒏𝜶 ≈ 𝝀

h

……(3)

According to the equation (3), product of two uncertainities lies in the range of ‘h’. If the calculation is done more carefully, then we come to know that, ∆X . ∆Px ≥

𝒉 𝟒𝝅

This equation indicates that greater the accuracy in determining the position, greater the uncertainity in determining the momentum. Thus a certainity in one quantity introduces an uncertainity in its conjugate quantity. In other words, if one quantity is known free from error, then the error in the other quantity becomes infinity. When

∆X = 0 ∆P =

𝒉 𝟒𝝅 𝐱 ∆𝐗

=∞

Uncertainity of velocity: According to the definition of momentum, ∆P = m x ∆v It means that, ∆X = m x ∆v ≥ ∆v x ∆X ≥

𝒉 𝟒𝝅𝒎

𝒉 𝟒𝝅

So, it is difficult to determine the velocity and position of electron simultaneously. Uncertainity principle can also be applied for another conjugate pair i.e. energy and time. Since, ∆𝜐 =

𝟏 ∆𝐭

As,

∆𝐄 = h x ∆𝜐

So,

∆𝐄 =

𝒉 ∆𝐭

∆𝐄 x ∆𝐭 = h More realistic treatment shows that, ∆𝐄 x ∆𝐭 =

𝒉 𝟐𝝅

It means that it is difficult to determine the energy and time for the particle simultaneously. So Heisenberg uncertainity principle is applicable to any conjugate pairs of variable and we reach the conclusion that the product of uncertainities of any two conjugate variable is always constant and its value range between

𝒉 𝟐𝝅

and

𝒉

.

𝟒𝝅

Uncertainity principle and particle of different sizes: Consider the motion of electron in hydrogen atom. Suppose that the position of electron from the nucleus can be determined with an uncertainity of 4 pm i.e. 4 x 10-12m. This means that the revolving electron may lie in first Bohr’s orbit having the radius in the range of 53 ± 4. (53 pm is the radius of first bohr’s orbit). The corresponding uncertainity is defining the velocity will be, ∆v = ∆v =

𝐡 𝟐𝝅𝐦∆𝐗 𝟔.𝟔𝟐𝟓 𝐱 𝟏𝟎−𝟑𝟒 𝐉𝐬 𝟐 x 𝟑.𝟏𝟒 x 𝟗.𝟏 x 𝟏𝟎−𝟑𝟏 𝐤𝐠 x 𝟒 x 𝟏𝟎−𝟏𝟐

∆v = 2.9 x 107m.sec-1 This is the uncertainity of the velocity, in Bohr’s first orbit. The velocity of electron in first Bohr’s orbit comes out to 2.2 x 106m.sec-1. The value of ∆v is much large than the velocity of electron. It means that the trajectory of electron cannot be defined. It means that Bohr’s orbits become meaningless in the light of uncertainity principle. Let us consider a ball of mass 2 g and having an uncertainity inn its position ∆X as 10-10m. The uncertainity in its velocity will be ∆v = ∆v =

𝐡 𝟐𝝅𝐦∆𝐗 𝟔.𝟔𝟐𝟓 𝐱 𝟏𝟎−𝟑𝟒 𝐉𝐬 𝟐 x 𝟑.𝟏𝟒 x 𝟐 x 𝟏𝟎−𝟑 𝐤𝐠 x 𝟏𝟎−𝟏𝟎 𝐦

∆v = 0.525 x 10-21m.sec-1 This uncertainity of velocity is negligible as compared to the velocity of a ball. Hence, the both velocity position of macroscopic particle can be determined precisely.

Applications of Uncertainity Principle: a) Electron cannot exit in the nucleus: We know that the radius of the nucleus of the atom is in the range of 10-14m. For an electron to remain whithin the nucleus, the “∆X”, the value of ‘∆v’ will be calculated as follows: ∆v ≥ ∆v ≥

𝐡 𝟒𝝅𝐦∆𝐗 𝟔.𝟔𝟐𝟓 𝐱 𝟏𝟎−𝟑𝟒 𝐤𝐠.𝐦𝟐 𝒔𝒆𝒄−𝟏 𝟒 x 𝟑.𝟏𝟒𝟐 x 𝟗.𝟏𝟎𝟖 x 𝟏𝟎−𝟑𝟏 𝐤𝐠 x 𝟏𝟎−𝟏𝟒 𝐦

∆v ≥ 5.77 x 109m.s-1 This value of ∆v is greater than the velocity of light i.e. 3 x 108m.sec-1. So an electron cannot exist in the nucleus. b) Probability concept of electronic cloud: We can never locate the exact position around the nucleus for the revolving electron. Example: The uncertainity in the momentum of a particle is 3.5 x 10-2kg ms-1. Calculate the uncertainity in its position. Solution:

Uncertainity in momentum

∆P = 3.5 x 10-2kg m.s-1 h = 6.625 x 10-34Js = 6.625 x 10-342kg m.s-1 Accorading to Heisenberg’s uncertainity principle, ∆X x ∆P ≥ ∆X x ≥

ℎ 4𝜋

ℎ ∆P x 4𝜋

Putting values ∆x ≥

𝟔.𝟔𝟐𝟓 𝐱 𝟏𝟎−𝟑𝟒 𝐤𝐠.𝐦𝟐 𝒔𝒆𝒄−𝟏 𝟑.𝟓 x 𝟏𝟎−𝟐 𝐤𝐠 𝒎𝒔−𝟏 x 4 x 3.14

∆x ≥ 0.1507 x 10-32m = 1.507 x 10-23nm