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Chapter 17: Problem 2398

The uncertainty in position of a particle is same as its de-Broglie wavelength, uncertainty in its momentum is \(\ldots \ldots\) (A) \((\overline{\mathrm{h}} / \lambda)\) (B) \((2 \overline{\mathrm{h}} / 3 \lambda)\) (C) \((\lambda \lambda) \overline{\mathrm{h}}\) (D) \((3 N 2 \lambda) \overline{\mathrm{h}}\)

Short Answer

Expert verified
The uncertainty in momentum is approximately \(\Delta p \geq \frac{2\hbar}{3\lambda}\). Option (B) is the correct answer.

Step by step solution

01

Write down the Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle states that: \[ \Delta x \Delta p \geq \frac{\hbar}{2} \] Where - \(\Delta x\) is the uncertainty in position, - \(\Delta p\) is the uncertainty in momentum, and - \(\hbar\) is the reduced Planck constant (\(\hbar = \frac{h}{2\pi}\)).
02

Write down the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[ \lambda = \frac{h}{p} \] Where - \(\lambda\) is the wavelength, - \(h\) is the Planck constant, and - \(p\) is the momentum.
03

Find the uncertainty in position

We are given that the uncertainty in position (\(\Delta x\)) is the same as the de Broglie wavelength (\(\lambda\)): \[ \Delta x = \lambda \]
04

Substitute the uncertainty in position into the Heisenberg Uncertainty Principle

Now, we can substitute the value of \(\Delta x\) into the Heisenberg Uncertainty Principle equation: \[ \lambda \Delta p \geq \frac{\hbar}{2} \] Next, we'll solve for the uncertainty in momentum (\(\Delta p\)).
05

Solve for the uncertainty in momentum

To solve for \(\Delta p\), we'll divide both sides of the inequality by \(\lambda\): \[ \Delta p \geq \frac{\hbar}{2\lambda} \] Now we have an inequality for the uncertainty in momentum.
06

Compare the inequality with the given options

We need to compare the inequality we obtained for the uncertainty in momentum (\(\Delta p \geq \frac{\hbar}{2\lambda}\)) with the given options: (A) \(\frac{\hbar}{\lambda}\) (B) \(\frac{2\hbar}{3\lambda}\) (C) \(\lambda\hbar\) (D) \(3N2\lambda\hbar\) The option closest to our result is (B) \(\frac{2\hbar}{3\lambda}\). Therefore, the uncertainty in momentum is approximately \(\frac{2\hbar}{3\lambda}\).

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Most popular questions from this chapter

With how much p.d. should an electron be accelerated, so that its de-Broglie wavelength is \(0.4 \AA\) (A) \(9410 \mathrm{~V}\) (B) \(94.10 \mathrm{~V}\) (C) \(9.140 \mathrm{~V}\) (D) \(941.0 \mathrm{~V}\)

The de-Broglie wavelength associated with a particle with rest mass \(\mathrm{m}_{0}\) and moving with speed of light in vacuum is..... (A) \(\left(\mathrm{h} / \mathrm{m}_{0} \mathrm{c}\right)\) (B) 0 (C) \(\infty\) (D) \(\left(\mathrm{m}_{0} \mathrm{c} / \mathrm{h}\right)\)

Calculate the energy of a photon of radian wavelength \(6000 \AA\) in \(\mathrm{eV}\) (A) \(20.6 \mathrm{eV}\) (B) \(2.06 \mathrm{eV}\) (C) \(1.03 \mathrm{eV}\) (D) \(4.12 \mathrm{eV}\)

Radius of a beam of radiation of wavelength \(5000 \AA\) is $10^{-3} \mathrm{~m}\(. Power of the beam is \)10^{-3} \mathrm{~W}$. This beam is normally incident on a metal of work function \(1.9 \mathrm{eV}\). The charge emitted by the metal per unit area in unit time is \(\ldots \ldots \ldots\) Assume that each incident photon emits one electron. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(1.282 \mathrm{C}\) (B) \(12.82 \mathrm{C}\) (C) \(128.2 \mathrm{C}\) (D) \(1282 \mathrm{C}\)

An electron is accelerated between two points having potential $20 \mathrm{~V}\( and \)40 \mathrm{~V}$, de-Broglic wavelength of electron is \(\ldots \ldots\) (A) \(0.75 \AA\) (B) \(7.5 \AA\) (C) \(2.75 \AA\) (D) \(0.75 \mathrm{~nm}\)
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