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Chapter 17: Problem 2357

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

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

Step by step solution

01

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.
02

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).
03

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.
04

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.
05

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.

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Most popular questions from this chapter

Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)

Find the velocity at which mass of a proton becomes \(1.1\) times its rest mass, \(\mathrm{m}_{\mathrm{g}}=1.6 \times 10^{-27} \mathrm{~kg}\) Also, calculate corresponding temperature. For simplicity, consider a proton as non- interacting ideal-gas particle at \(1 \mathrm{~atm}\) pressure. $\left(1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J} \cdot \mathrm{h}=6.63 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right)$ (A) $\mathrm{V}=1.28 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{12} \mathrm{~K}$ (B) $\mathrm{V}=12.6 \times 10^{8}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.65 \times 10^{11} \mathrm{~K}$ (C) $\mathrm{V}=1.26 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=5.76 \times 10^{11} \mathrm{~K}$ (D) $\mathrm{V}=12.6 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.56 \times 10^{11} \mathrm{~K}$

In photo electric effect, if threshold wave length of a metal is \(5000 \AA\) work function of this metal is ..........V. $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(1.24\) (B) \(2.48\) (C) \(4.96\) (D) \(3.72\)

An electric bulb of \(100 \mathrm{w}\) converts \(3 \%\) of electrical energy into light energy. If the wavelength of light emitted is \(6625 \AA\), the number of photons emitted is \(1 \mathrm{~s}\) is \(\ldots \ldots\) $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(10^{17}\) (B) \(10^{19}\) (C) \(10^{21}\) (D) \(10^{15}\)

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