
According to de Broglie, which of the following statements is true about the wavelength of a moving particle?
(a) It is never large enough to measure.
(b) It is proportional to the speed of the particle.
(c) It is inversely proportional to the momentum of the particle.
(d) it is equal to Planck’s constant
(e) it does not affect the behaviour of electrons.
Answer
224.1k+ views
Hint: According to the question one must know about the concept of de Broglie wavelength associated with the particle. And then only one can solve this question. When studying quantum mechanics, the de Broglie wavelength is a key idea. De Broglie wavelength is the wavelength () that is connected to an item concerning its momentum and mass. Typically, a particle's force is inversely proportional to its de Broglie wavelength.
Formula used: ${\lambda _{dB}} = \dfrac{h}{p}$
Where h is a Planck’s constant.
p is momentum and
${\lambda _{dB}}$ is de Broglie’s wavelength
Complete Step by Step Solution:
As we know that the de Broglie wavelength concept is associated with the particle i.e., ${\lambda _{dB}} = \dfrac{h}{p}$,
From here we can see that de Broglie’s wavelength is inversely proportional to the momentum and directly proportional to Planck’s constant. This means as the momentum will increase the value of de Broglie wavelength will decrease.
Therefore the correct answer is option (c).
Note: Note that the S.I unit of de Broglie wavelength is meter (m). By analysing the diffraction pattern created as electrons flow through a crystalline substance, we can deduce that matter has a wave-like character. The pattern appears when the electrons' de Broglie wavelength and the distance between their atoms in the crystals are similar.
Formula used: ${\lambda _{dB}} = \dfrac{h}{p}$
Where h is a Planck’s constant.
p is momentum and
${\lambda _{dB}}$ is de Broglie’s wavelength
Complete Step by Step Solution:
As we know that the de Broglie wavelength concept is associated with the particle i.e., ${\lambda _{dB}} = \dfrac{h}{p}$,
From here we can see that de Broglie’s wavelength is inversely proportional to the momentum and directly proportional to Planck’s constant. This means as the momentum will increase the value of de Broglie wavelength will decrease.
Therefore the correct answer is option (c).
Note: Note that the S.I unit of de Broglie wavelength is meter (m). By analysing the diffraction pattern created as electrons flow through a crystalline substance, we can deduce that matter has a wave-like character. The pattern appears when the electrons' de Broglie wavelength and the distance between their atoms in the crystals are similar.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

How to Convert a Galvanometer into an Ammeter or Voltmeter

Hybridisation in Chemistry – Concept, Types & Applications

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

