
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
216.3k+ 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.
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