Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 2329

At \(10^{\circ} \mathrm{C}\) temperature, de-Broglie wave length of atom is $0.4 \AA\(.If temperature of atom is increased by \)30^{\circ} \mathrm{C}$, what will be change in de-Broglie wavelength of atom? (A) decreases \(10^{-2} \AA\) (B) decreases \(2 \times 10^{-2} \AA\) (C) increases \(10^{-2} \AA\) (D) increases \(2 \times 10^{-2} \AA\)

Short Answer

Expert verified
The change in de-Broglie wavelength, \(\Delta\lambda\), is equal to an increase of \(2 \times 10^{-2} \AA.\) The correct answer is (D).

Step by step solution

01

Write the de-Broglie wavelength formula for an atom

The de-Broglie wavelength formula for an atom can be given as follows: \( \lambda = \frac{h}{p} \) where \(\lambda\) is the de-Broglie wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the atom.
02

Write the temperature-energy relation for an atom

The temperature has an impact on the kinetic energy of an atom, which in turn impacts its momentum. The average kinetic energy of an atom can be expressed in terms of temperature as follows: \( \text{KE} = \frac{3}{2}k_\mathrm{B}T \) where \(\text{KE}\) is the average kinetic energy of an atom, \(k_\mathrm{B}\) is the Boltzmann constant, and \(T\) is the temperature in Kelvin.
03

Calculate the initial momentum and wavelength of the atom

We are given that the initial de-Broglie wavelength of the atom is \(0.4\, \text{\AA}\) or \(4 \times 10^{-11}\, \text{m}\) at \(10^{\circ} \mathrm{C}\) or \(283.15\, \text{K}\). We can use the de-Broglie wavelength formula and temperature-energy relation to find the initial momentum of the atom. From the de-Broglie wavelength formula, we can get the momentum as: \( p = \frac{h}{\lambda} \) Since the average kinetic energy of an atom is related to temperature as mentioned in Step 2, we can use that relation to find the initial momentum: \( p^2 = 2m(\text{KE}) \)
04

Calculate the final momentum and wavelength of the atom

We are given that the temperature is increased by \(30^{\circ} \mathrm{C}\), i.e., the final temperature is \((10 + 30)^{\circ} \mathrm{C}\) or \(313.15\, \text{K}\). We can use the temperature-energy relation to find the final momentum of the atom and then use the de-Broglie wavelength formula to find the final wavelength of the atom.
05

Calculate the change in de-Broglie wavelength

After finding the initial and final wavelengths of the atom, we can calculate the change in de-Broglie wavelength by subtracting the initial wavelength from the final wavelength: \( \Delta\lambda = \lambda_\mathrm{final} - \lambda_\mathrm{initial} \) By calculating the value of \(\Delta\lambda\), we can determine which option among (A), (B), (C), and (D) is correct, considering that each option states whether the change in de-Broglie wavelength results in an increase or decrease and by what magnitude.

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

Photoelectric effect is obtained on metal surface for a light having frequencies \(\mathrm{f}_{1} \& \mathrm{f}_{2}\) where \(\mathrm{f}_{1}>\mathrm{f}_{2}\). If ratio of maximum kinetic energy of emitted photo electrons is \(1: \mathrm{K}\), so threshold frequency for metal surface is \(\ldots \ldots \ldots \ldots\) (A) $\left\\{\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (B) $\left\\{\left(\mathrm{Kf}_{1}-\mathrm{f}_{2}\right) /(\mathrm{K}-1)\right\\}$ (C) $\left\\{\left(\mathrm{K} \mathrm{f}_{2}-\mathrm{f}_{1}\right) /(\mathrm{K}-1)\right\\}$ (D) \(\left\\{\left(\mathrm{f}_{2}-\mathrm{f}_{1}\right) / \mathrm{K}\right\\}\)

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) \(3.28 \mathrm{~V}\)

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) \(I-R . I I-S . I I I-P . I V-O\)

Kinetic energy of proton accelerated under p.d. \(1 \mathrm{~V}\) will be........ (A) \(1840 \mathrm{eV}\) (B) \(13.6 \mathrm{eV}\) (C) \(1 \mathrm{eV}\) (D) \(0.54 \mathrm{eV}\)

A star which can be seen with naked eye from Earth has intensity $1.6 \times 10^{-9} \mathrm{Wm}^{-2}\( on Earth. If the corresponding wavelength is \)560 \mathrm{~nm}\(, and the radius of the human eye is \)2.5 \times 10^{-3} \mathrm{~m}\(, the number of photons entering in our in \)1 \mathrm{~s}$ is..... (A) \(7.8 \times 10^{4} \mathrm{~s}^{-1}\) (B) \(8.85 \times 10^{4} \mathrm{~s}^{-1}\) (C) \(7.85 \times 10^{5} \mathrm{~s}^{-1}\) (D) \(8.85 \times 10^{5} \mathrm{~s}^{-1}\)
See all solutions

Recommended explanations on English Textbooks

Language Analysis

Read Explanation

5 Paragraph Essay

Read Explanation

Discourse

Read Explanation

Graphology

Read Explanation

Key Concepts in Language and Linguistics

Read Explanation

Text Comparison

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