De Broglie wavelength $\lambda $ associated with neutrons is related with absolute temperature $T$ as:
(A) $\lambda \propto T$
(B) $\lambda \propto \dfrac{1}{T}$
(C) $\lambda \propto \dfrac{1}{{\sqrt T }}$
(D) $\lambda \propto {T^2}$
Answer
299.4k+ views
Hint De Broglie proposed that matter is associated with a wave called matter-wave. According to him matter and energy could have symmetrical characteristics. The radiant energy has dual characteristics. Hence the matter should exhibit dual characteristics. We can find the temperature using the expression for the De Broglie wavelength of a particle.
Formula used:
$\lambda = \dfrac{h}{{mv}}$
Where, $\lambda $ stands for the De Broglie wavelength of the particle, $h$ is the Planck’s constant, $m$ stands for the mass of the particle, $v$ stands for the velocity of the particle.
Complete step by step answer:
Consider a particle having a mass, $m$ moving with a velocity $v$.
The De Broglie wavelength of the matter wave of the particle is given by,
$\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}$
Where $p$ is the momentum of the particle.
The kinetic energy of the particle can be written as
$E = \dfrac{1}{2}m{v^2}$
From this equation we get
${v^2} = \dfrac{{2E}}{m} $
$\Rightarrow v = \sqrt {\dfrac{{2E}}{m}}$
Substituting this value of $v$in the expression for wavelength, we get
$\lambda = \dfrac{h}{{m\sqrt {\dfrac{{2E}}{m}} }} = \dfrac{h}{{\sqrt {2mE} }}$
We know that the energy of neutrons is directly proportional to the temperature, i.e. $E \propto T$
Since, $\lambda \propto \dfrac{1}{{\sqrt E }}$ and $E \propto T$
We can write that
$\lambda \propto \dfrac{1}{{\sqrt T }}$
The answer is Option (C): $\lambda \propto \dfrac{1}{{\sqrt T }}$
Note
All the particles in quantum mechanics are manifested with De Broglie wavelength. The De Broglie wavelength gives the probability of finding a particle in a given configuration space. The equation for the De Broglie wavelength gives the wavelength of matter-wave. The wavelength is appreciable when the mass is very small and the velocity is very large. The matter-wave associated with moving electrons can be verified by crystal diffraction experiments (Davisson and Germer experiment). De Broglie was awarded the Nobel Prize for his discovery of the wave nature of electrons in 1929.
Formula used:
$\lambda = \dfrac{h}{{mv}}$
Where, $\lambda $ stands for the De Broglie wavelength of the particle, $h$ is the Planck’s constant, $m$ stands for the mass of the particle, $v$ stands for the velocity of the particle.
Complete step by step answer:
Consider a particle having a mass, $m$ moving with a velocity $v$.
The De Broglie wavelength of the matter wave of the particle is given by,
$\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}$
Where $p$ is the momentum of the particle.
The kinetic energy of the particle can be written as
$E = \dfrac{1}{2}m{v^2}$
From this equation we get
${v^2} = \dfrac{{2E}}{m} $
$\Rightarrow v = \sqrt {\dfrac{{2E}}{m}}$
Substituting this value of $v$in the expression for wavelength, we get
$\lambda = \dfrac{h}{{m\sqrt {\dfrac{{2E}}{m}} }} = \dfrac{h}{{\sqrt {2mE} }}$
We know that the energy of neutrons is directly proportional to the temperature, i.e. $E \propto T$
Since, $\lambda \propto \dfrac{1}{{\sqrt E }}$ and $E \propto T$
We can write that
$\lambda \propto \dfrac{1}{{\sqrt T }}$
The answer is Option (C): $\lambda \propto \dfrac{1}{{\sqrt T }}$
Note
All the particles in quantum mechanics are manifested with De Broglie wavelength. The De Broglie wavelength gives the probability of finding a particle in a given configuration space. The equation for the De Broglie wavelength gives the wavelength of matter-wave. The wavelength is appreciable when the mass is very small and the velocity is very large. The matter-wave associated with moving electrons can be verified by crystal diffraction experiments (Davisson and Germer experiment). De Broglie was awarded the Nobel Prize for his discovery of the wave nature of electrons in 1929.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

Hybridisation in Chemistry – Concept, Types & Applications

Isoelectronic Species: Definition, Examples & Importance

