Spread the love

State Heisenberg Uncertainty Principle

State Heisenberg Uncertainty Principle :- According to Heisenberg it is impossible to measure the position and momentum of a particle simultaneously with 100% accuracy. This is called Heisenberg Uncertainty Principle.

If Δx and Δp are the uncertainties in determining the position and momentum of the particle simultaneously, then

$\displaystyle \Delta x\Delta p\ge \frac{h}{4\pi }$ …..(1)

here h = 6.63 × 10^-34Js is Plank’s constant

From equation (1) if we are able to measure the exact position of the particle(Δx = 0) then the uncertainty in the measurement of its linear momentum is infinite(Δp→∞). In the same way if we are able to measure the exact linear momentum of the particle(Δp = 0) then we cannot measure the exact position of the particle at that time.

∴ Δx = 0, then Δp→∞ and if Δp = 0 then Δx→∞

If we have to observe an electron then it cannot be done without light and when light(photons) falls on electron, then it’s momentum changes as shown in figure below :-

Uncertainty Principle is a fundamental part of nature and it holds for all microscopic and macroscopic particles.

Other forms of Heisenberg Uncertainty Principle

(State Heisenberg Uncertainty Principle)

(a) In terms of position and velocity :-

As momentum p = mv, Δp = mΔv, so from equation (1), we get

$\displaystyle m\Delta x\Delta \text{v}\ge \frac{h}{4\pi }$

$\displaystyle \Rightarrow \Delta x\Delta \text{v}\ge \frac{h}{4\pi m}$ …..(2)

(b) In terms of energy and time :-

Now by dividing and multiplying equation (1) by Δt, we get

$\displaystyle \left( \frac{\Delta x}{\Delta t}\Delta p \right)\Delta t\ge \frac{h}{4\pi }$

As Δp = FΔt, so

$\displaystyle \Rightarrow \left( \frac{\Delta x}{\Delta t}F\Delta t \right)\Delta t\ge \frac{h}{4\pi }$

$\displaystyle \Rightarrow \left( F\Delta x \right)\Delta t\ge \frac{h}{4\pi }$

As FΔx = ΔE, hence

$\displaystyle \Delta E\Delta t\ge \frac{h}{4\pi }$ …..(3)

(c) In terms of angular momentum and angular displacement :-

If the trajectory of the particle is circular with radius $r$

$\displaystyle \Delta L=r\Delta p$

$\displaystyle \Delta \theta =\frac{\Delta l}{r}$

where Δl is the uncertainty in the measurement of position along the trajectory.

Thus

$\displaystyle \Delta L\Delta \theta =r\Delta p\frac{\Delta l}{r}=\Delta p\Delta l\ge \frac{h}{4\pi }$

$\displaystyle \Rightarrow \Delta L\Delta \theta \ge \frac{h}{4\pi }$ …..(4)

Example 1

If uncertainty in position and momentum are equal, then uncertainty in velocity is ?

(A) $\displaystyle \sqrt{\frac{h}{\pi }}$

(B) $\displaystyle \frac{1}{2m}\sqrt{\frac{h}{\pi }}$

(D) $\displaystyle \frac{1}{m}\sqrt{\frac{h}{\pi }}$

Solution :-

According to Heisenberg uncertainty principle,

$\displaystyle \Delta x\Delta p=\frac{h}{4\pi }$

Given Δx = Δp

Hence,

$\displaystyle {{\left( \Delta p \right)}^{2}}=\frac{h}{4\pi }$

$\displaystyle \Rightarrow \Delta p=\sqrt{\frac{h}{4\pi }}$

$\displaystyle \Rightarrow m\Delta v=\frac{1}{2}\sqrt{\frac{h}{\pi }}$

$\displaystyle \therefore \Delta v=\frac{1}{2m}\sqrt{\frac{h}{\pi }}$

Hence correct option is (B).

Example 2

If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron.

Solution :-

Here

Δx = 0.002 nm = 0.002 × 10 ^-9m = 2 × 10 ^-12m

Using Heisenberg uncertainty principle,

$\displaystyle \Delta x\Delta p=\frac{h}{4\pi }$

$\displaystyle \Rightarrow \Delta p=\frac{1}{\Delta x}\frac{h}{4\pi }=\frac{1}{\left( 2\times {{10}^{-12}} \right)}\times \frac{6.63\times {{10}^{-34}}}{4\times 3.14}$

$\displaystyle \therefore \Delta p=2.638\times {{10}^{-23}}kgm{{s}^{-1}}$

Example 3

The measurement of the electron position is associated with an uncertainty in momentum, which is equal to $1 \times 10^{- 18} g c m s^{- 1}$ . The uncertainty in electron velocity is (mass of an electron is $9 \times 10^{- 28} g$ )