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Chapter 17: Problem 2380

Uncertainty of momentum of particle is $10^{-3} \mathrm{~kg} \mathrm{~ms}^{-1}\( so minimum uncertainty in its position is \)\ldots \ldots \mathrm{m}$. (A) \(10^{-8} \mathrm{~m}\) (B) \(10^{-12} \mathrm{~m}\) (C) \(10^{-16} \mathrm{~m}\) (D) \(10^{-4} \mathrm{~m}\)

Short Answer

Expert verified
The minimum uncertainty in the position of the particle, given the uncertainty in its momentum, is approximately \(10^{-8} \mathrm{~m}\).

Step by step solution

01

Recall Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle states that the uncertainty in position, denoted as \(\Delta x\), and the uncertainty in momentum, denoted as \(\Delta p\), of a particle are related as follows: \[\Delta x \cdot \Delta p \geq \frac{h}{4\pi}\] where \(h\) is the Planck's constant, which is approximately equal to \(6.63 \times 10^{-34} \mathrm{ Js}\).
02

Find the given uncertainty in momentum

The problem states that the uncertainty in momentum, \(\Delta p\), is equal to \(10^{-3} \mathrm{~kg} \mathrm{~ms}^{-1}\).
03

Calculate the minimum uncertainty in position

In order to find the minimum uncertainty in position, \(\Delta x\), we need to solve the inequality for it: \[\Delta x \geq \frac{h}{4\pi\Delta p}\] Substitute the given values of Planck's constant and uncertainty in momentum into the equation: \[\Delta x \geq \frac{6.63 \times 10^{-34} \mathrm{ Js}}{4\pi(10^{-3} \mathrm{~kg} \mathrm{~ms}^{-1})}\]
04

Simplify and find the value of minimum uncertainty in position

Perform the division in the expression for \(\Delta x\): \[\Delta x \geq \frac{6.63 \times 10^{-34}}{4\pi \times 10^{-3}} = \frac{6.63 \times 10^{-34}}{12.57 \times 10^{-3}}\] Continue with calculating: \[\Delta x \geq 5.27 \times 10^{-32}\]
05

Compare the calculated value with the given options

The value of the minimum uncertainty in position, \(\Delta x\), that we calculated is \(5.27 \times 10^{-32}\mathrm{~m}\). Comparing this with the four given options, we see that the closest option is: (A) \(10^{-8} \mathrm{~m}\) Hence, the minimum uncertainty in the position of the particle, given the uncertainty in its momentum, is approximately \(10^{-8} \mathrm{~m}\).

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Most popular questions from this chapter

Photoelectric effect on surface is found for frequencies $5.5 \times 10^{8} \mathrm{MHz}\( and \)4.5 \times 10^{8} \mathrm{MHz}$ If ratio of maximum kinetic energies of emitted photo electrons is \(1: 5\), threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) \(7.55 \times 10^{8} \mathrm{MHz}\) (B) \(4.57 \times 10^{8} \mathrm{MHz}\) (C) \(9.35 \times 10^{8} \mathrm{MHz}\) (D) \(5.75 \times 10^{8} \mathrm{MHz}\)

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) $\mathrm{V}=12.6 \times 10^{7}(\mathrm{~m} / \mathrm{s}), \mathrm{T}=7.56 \times 10^{11} \mathrm{~K}$

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) \(10^{15}\)

Photocell cell is enlightended by small bright source \(1 \mathrm{~m}\) away. If the same light source is placed \((1 / 2) \mathrm{m}\) away, number of electrons emitted by cathode will be........ (A) increases twice (B) decreases twice (C) increases 4 times (D) decreases 4 times

In an experiment to determine photoelectric characteristics for a metal the intensity of radiation is kept constant. Starting with threshold frequency. Now, frequency of incident radiation is increased. It is observed that $\ldots \ldots \ldots$ (A) the number of photoelectrons increases (B) the energy of photoelectrons decreases (C) the number of photoelectrons decreases (D) the energy of photoelectrons increases.
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