Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2379

de-Broglie wavelength of atom at T \(\mathrm{K}\) absolute temperature will be \(\ldots \ldots \ldots\) (A) \([\mathrm{h} /\\{\mathrm{mkT}\\}]\) (B) \([\mathrm{h} /\\{\sqrt{3} \mathrm{mKT}\\}]\) (C) \([\\{\sqrt{2} \mathrm{mKT}\\} / \mathrm{h}]\) (D) \(\sqrt{(2 \mathrm{mKT})}\)

Short Answer

Expert verified
The short answer for the de-Broglie wavelength of an atom at T K absolute temperature is given by: \[λ = \left[\frac{h}{\sqrt{3mKT}}\right]\] where λ is the de-Broglie wavelength, h is the Planck's constant, m is the mass of the atom, K is the Boltzmann constant, and T is the temperature in Kelvin. So, the correct option is (B).

Step by step solution

01

Calculate the momentum of the atom

Let's start by finding the momentum of the atom with a given kinetic energy. The relation between momentum (p) and kinetic energy (K.E.) is given by the formula: \[K.E. = \frac{p^2}{2m}\] Now, let's plug in the kinetic energy in terms of temperature: \[\frac{3}{2}kT = \frac{p^2}{2m}\]
02

Solve for the momentum (p)

Next, let's solve for the momentum (p) in the above equation: \(p^2 = 3mkT\)
03

Calculate the de-Broglie wavelength (λ)

Now that we have the expression for momentum, let's use the de-Broglie wavelength formula: \[λ = \frac{h}{p}\] Plug in the expression for momentum: \[λ = \frac{h}{\sqrt{3mkT}}\] The above expression matches the option (B) in the given exercise: \[λ = \left[\frac{h}{\sqrt{3mKT}}\right]\] Therefore, the correct answer is (B).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Matching type questions: (Match, Column-I and Column-II property) Column-I Column-II (I) Quantization of charge (P) Diffraction of light (II) Wave nature of light (Q) de Broglie hypothesis (III) Dual nature of matter (R) Photo-electric effect (IV) Particle nature of light (S) Millikan's drop experiment (A) $\mathrm{I}-\mathrm{P}, \mathrm{II}-\mathrm{Q}, \mathrm{III}-\mathrm{R}, \mathrm{IV}-\mathrm{S}$ (B) $\mathrm{I}-\mathrm{S}, \mathrm{II}-\mathrm{P}, \mathrm{III}-\mathrm{Q}, \mathrm{IV}-\mathrm{R}$ (C) $\mathrm{I}-\mathrm{Q}, \mathrm{II}-\mathrm{R}, \mathrm{III}-\mathrm{S}, \mathrm{IV}-\mathrm{P}$ (D) \(I-R . I I-S . I I I-P . I V-O\)

Consider the radius of a nucleus to be \(10^{-15} \mathrm{~m} .\) If an electron is assumed to be in such nucleus, what ill be its energy? $\left(\mathrm{me}=9.1 \times 10^{-31} \mathrm{~kg} \cdot \mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}\right)$ (A) \(5.59 \times 10^{3} \mathrm{MeV}\) (B) \(9.55 \times 10^{3} \mathrm{MeV}\) (C) \(5.95 \times 10^{3} \mathrm{MeV}\) (D) \(7.45 \times 10^{3} \mathrm{MeV}\)

De-Broglie wavelength of particle moving at a (1/4) th of speed of light having rest mass \(\mathrm{m}_{0}\) is \(\ldots \ldots \ldots\) (A) $\left\\{(3.87 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (B) $\left\\{(4.92 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (C) $\left\\{(7.57 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (D) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

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) \(1282 \mathrm{C}\)

Compare energy of a photon of X-rays having \(1 \AA\) wavelength with the energy of an electron having same de Broglie wavelength. $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, 1 \mathrm{ev}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(8.24\) (B) \(2.48\) (C) \(82.4\) (D) \(24.8\)
See all solutions

Recommended explanations on English Textbooks

Syntax

Read Explanation

Single Paragraph Essay

Read Explanation

Argumentative Essay

Read Explanation

Rhetoric

Read Explanation

Phonetics

Read Explanation

International English

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free