Question

Uncertainty in position of electron is found of the order of de-Broglie wavelength. Using Heisenberg's uncertainty principle, it is found that order of uncertainty in its velocity \(=\ldots \ldots \ldots .\) (A) \(1 \mathrm{v}\) (B) \(2 \mathrm{v}\) (C) \((\mathrm{v} / 2 \pi)\) (D) \(2 \pi \mathrm{v}\)

Short answer

The order of uncertainty in an electron's velocity is 1v (option A).

Step-by-step solution

Heisenberg's Uncertainty Principle

The Heisenberg's uncertainty principle states that the product of the uncertainties in position (Δx) and momentum (Δp) is always greater than or equal to a constant value, which is given as: \[\Delta x \cdot \Delta p \geq \frac{\hbar}{2}\] where; Δx = uncertainty in position (of the order of its de-Broglie wavelength) Δp = uncertainty in momentum (mass × uncertainty in velocity) m = mass of the electron Δv = uncertainty in velocity h = Planck's constant \(\hbar = \frac{h}{2 \pi}\) = reduced Planck's constant (h-bar) Since we are given the uncertainty in position is of the order of its de-Broglie wavelength, let's first calculate the de-Broglie wavelength.

Calculate the de-Broglie wavelength

The de-Broglie wavelength (λ) is given by: \[\lambda = \frac{h}{p}\] where; λ = de-Broglie wavelength h = Planck's constant p = momentum (mass × velocity) Since we are working in terms of uncertainty, the uncertainty in position (Δx) will be of the order of de-Broglie wavelength (λ). Therefore, \[\Delta x = \lambda\] Now, let's substitute Δx with λ and find the uncertainty in momentum (Δp).

Calculate the uncertainty in momentum

Using Heisenberg's uncertainty principle formula, the uncertainty in momentum (Δp) can be calculated as: \[\Delta p = \frac{\hbar}{2 \Delta x}\] Substitute Δx with λ: \[\Delta p = \frac{\hbar}{2 \lambda}\] Now, let's find the uncertainty in velocity (Δv) using the momentum formula.

Calculate the uncertainty in velocity

Since momentum (p) = mass(m) × velocity (v), the uncertainty in momentum (Δp) = mass (m) × uncertainty in velocity (Δv). Therefore, we can calculate the uncertainty in velocity (Δv) as: \[\Delta v = \frac{\Delta p}{m}\] Substitute Δp with the expression we found in step 3: \[\Delta v = \frac{\hbar}{2 \lambda m}\] Since we need the order of the uncertainty in its velocity, we can rewrite the above equation as: \[\Delta v \propto \frac{1}{\lambda}\] Now, let's compare this equation with the provided options to find the correct answer.

Match the correct option

Comparing our derived equation for the order of uncertainty in velocity (\(\Delta v \propto \frac{1}{\lambda}\)) with the given options, we find that the correct option is: (A) \(1 \mathrm{v}\) So, the order of the uncertainty in an electron's velocity is 1v.