Heisenberg Uncertainty Principle Formula and Application
If, ∆x is the error in position measurement and ∆p is the error in the measurement of momentum, then
∆X × ∆p ≥ \frac{h}{4\pi }
4π
h
Since momentum, p = mv, Heisenberg’s uncertainty principle formula can be alternatively written as-
∆X × ∆mv ≥ \frac{h}{4\pi }
4π
h
or ∆X × ∆m × ∆v ≥ \frac{h}{4\pi }
4π
h
Where, ∆V is the error in the measurement of velocity and assuming mass remaining constant during the experiment,
∆X × ∆V ≥ \frac{h}{4\pi m}
4πm
h
.
Accurate measurement of position or momentum automatically indicates larger uncertainty (error) in the measurement of the other quantity.
Applying the Heisenberg principle to an electron in an orbit of an atom, with h = 6.626 ×10-34Js and m= 9.11 ×10-31Kg,
∆X × ∆V ≥ \frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 9.11\times {{10}^{-31}}}
4×3.14×9.11×10
−31
6.626×10
−34
= 10-4 m2 s-1.
If the position of the electron is measured accurately to its size (10-10m), then the error in the measurement of its velocity will be equal or larger than 106m or 1000Km.
Heisenberg principle applies to only dual-natured microscopic particles and not to a macroscopic particle whose wave nature is very small.
Also Read: de Broglie Equation
Explaining Heisenberg Uncertainty Principle With An Example
Electromagnetic radiations and microscopic matter waves exhibit a dual nature of mass/ momentum and wave character. Position and velocity/momentum of macroscopic matter waves can be determined accurately, simultaneously. For example, the location and speed of a moving car can be determined at the same time, with minimum error. But, in microscopic particles, it will not be possible to fix the position and measure the velocity/momentum of the particle simultaneously.
An electron in an atom has a mass of 9.91 × 10-31Kg. Naked eyes will not see such small particles. A powerful light may collide with the electron and illuminate it. Illumination helps in identifying and measuring the position of the electron. The collision of the powerful light source, while helping in identification increases the momentum of the electron and makes it move away from the initial position. Thus, when fixing the position, velocity /momentum of the particle would have changed from the original value. Hence when the position is exact, error occurs in the measurement of velocity or momentum. In the same way, the measurement of momentum accurately will change the position.
Hence, at any point in time, either position or momentum can only be measured accurately.
Simultaneous measurement of both of them will have an error in both position and momentum. Heisenberg quantified the error in the measurement of both position and momentum at the same time.
Heisenberg’s γ-ray Microscope
A striking thought experiment illustrating the uncertainty principle is Bohr’s / Heisenberg’s Gamma-ray microscope. To observe a particle, say an electron, we shine it with the light ray of wavelength λ and collect the Compton scattered light in a microscope objective whose diameter subtends an angle θ with the electron as shown in the figure below
Heisenberg’s γ-ray Microscope
The precision with which the electron can be located, Delta x, is defined by the resolving power of the microscope,
sin \theta =\frac{\lambda }{\Delta x}\Rightarrow \Delta x=\frac{\lambda }{sin \theta }sinθ=
Δx
λ
⇒Δx=
sinθ
λ
It appears that by making λ small, that is why we choose γ-ray, and by making sin θ large, Delta x can be made as small as desired. But, according to the uncertainty principle, we can do so only at the expense of our knowledge of x-component of electron momentum.
In order to record the Compton scattered photon by the microscope, the photon must stay in the cone of angle θ and hence its x-component of the momentum can vary within ±(h/λ) sin θ. This implies, the magnitude of the recoil momentum of the electron is uncertain by
\Delta p_{x}=\frac{2h}{\lambda }sin \thetaΔp
x
=
λ
2h
sinθ
The product of the uncertainty yields,
\Delta x\Delta p_{x}=\frac{\lambda }{sin \theta }\frac{2h}{\lambda }sin \theta =4\pi hΔxΔp
x
=
sinθ
λ
λ
2h
sinθ=4πh
Is Heisenberg’s Uncertainty Principle Noticeable in All Matter Waves?
Heisenberg’s principle is applicable to all matter waves. The measurement error of any two conjugate properties, whose dimensions happen to be joule sec, like position-momentum, time-energy will be guided by the Heisenberg’s value.
But, it will be noticeable and of significance only for small particles like an electron with very low mass. A bigger particle with heavy mass will show the error to be very small and negligible.
Heisenberg Uncertainty Principle Equations
Heisenberg’s uncertainty principle can be considered as a very precise mathematical statement that describes the nature of quantum systems. As such, we often consider two common equations related to the uncertainty principle. They are;
Equation 1: ∆X ⋅ ∆p ~ ħ
Equation 2: ∆E ⋅ ∆t ~ ħ
Where,
ħ = value of the Planck’s constant divided by 2*pi
∆X = uncertainty in the position
∆p = uncertainty in momentum
∆E = uncertainty in the energy
∆t = uncertainty in time measurement
Heisenberg Uncertainty Principle Problems With Solutions
1. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h / 4pm × 0.05 nm, is there any problem in defining this value.
a) ∆x = 2×10-12m; ∆X × ∆mV ≥ \frac{h}{4\pi }
4π
h
= \frac{6.626\times {{10}^{-34}}}{4\times 3.14}
4×3.14
6.626×10
−34
⸪ ∆mV ≥ \frac{h}{4\pi \Delta x}
4πΔx
h
≥ \frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 2\times {{10}^{-12}}}\,
4×3.14×2×10
−12
6.626×10
−34
= 2.64 × 10-23 Kg m s-1
b) Momentum mv = \,\frac{h\times 5\times {{10}^{-11}}}{4\times {{10}^{-12}}}=\frac{6.626\times {{10}^{-34}}\times 5\times {{10}^{-11}}}{4\times {{10}^{-12}}}\,\,
4×10
−12
h×5×10
−11
=
4×10
−12
6.626×10
−34
×5×10
−11
= 28 × 10-33
Error in momentum measurement is 1010 times larger than the actual momentum. The given momentum will not be acceptable.
2. Position of a chloride ion on a material can be determined to a maximum error of 1μm. If the mass of the chloride ion is 5.86 × 10-26Kg, what will be the error in its velocity measurement?
∆x = 10-6 m; ∆X × ∆mV ≥ \frac{h}{4\pi }
4π
h
= \,\frac{6.626\times {{10}^{-34}}}{4\times 3.14}
4×3.14
6.626×10
−34
= 5.28×10-35Js
⸪ ∆V ≥ \,\frac{h}{4\pi m\Delta x}\ge \frac{6.626\times {{10}^{-34}}}{4\times 3.14\times 5.86\times {{10}^{-26}}\times {{10}^{-6}}}\,
4πmΔx
h
≥
4×3.14×5.86×10
−26
×10
−6
6.626×10
−34
= 9 × 10-4m s-1
3. The lifetime of an excited state of an atom is 3 × 10-3s. What is the minimum uncertainty in its energy in eV?
Time and energy are conjugate pairs with Js unit. The product of measurement error is given by Heisenberg’s principle.
∆t × ∆E ≥ \frac{h}{4\pi }
4π
h
= \,\frac{6.626\times {{10}^{-34}}}{4\times 3.14}
4×3.14
6.626×10
−34
= 5.28×10-35Js
Assuming a maximum error in the measurement of lifetime equal to that of lifetime = 3 ×10-3s
∆E ≥ \,\frac{h}{4\pi m\Delta x}=\frac{1}{3\times {{10}^{-3}}}\,\,
4πmΔx
h
=
3×10
−3
1
× 5.28×10-35J
⸪ 1 Joule = 6.242 × 1018ev,
Uncertainty in the determination of energy of the atom = ∆E = 6.22 × 1018 × \frac{1}{3\times {{10}^{-3}}}\,
3×10
−3
1
× 5.28 ×10-35
= 1.1×10-13
