(C) The kinetic energy $K$ of a neutron in thermal equilibrium at temperature $T$ is given by the equipartition theorem as:$K = \frac{3}{2} k_B T$ ..

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(C) The kinetic energy $K$ of a neutron in thermal equilibrium at temperature $T$ is given by the equipartition theorem as:$K = \frac{3}{2} k_B T$ ..... $(i)$The relation between momentum $p$ and kinetic energy $K$ is $p = \sqrt{2mK}$.Substituting the value of $K$ from equation $(i)$:$p = \sqrt{2m \cdot \frac{3}{2} k_B T} = \sqrt{3mk_B T}$The de-Broglie wavelength $\lambda$ is given by:$\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mk_B T}}$

Class 12 Physics Dual Nature of Radiation and matter Matter Waves and de Broglie Wavelength

Medium

The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (kelvin) and mass $m$ is:

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A
$\frac{h}{\sqrt{mkBT}}$
B
$\frac{h}{\sqrt{2mkBT}}$
C
$\frac{h}{\sqrt{3mkBT}}$
D
$\frac{h}{2\sqrt{mkBT}}$

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Dual Nature of Radiation and matter

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Matter Waves and de Broglie Wavelength

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Which of the following graphs correctly represents the de-Broglie wavelength $\lambda$ associated with a fundamental particle having linear momentum $p$?

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An $\alpha$-particle moves in a circular path of radius $0.83\, cm$ in the presence of a magnetic field of $0.25\, Wb/m^2$. The de Broglie wavelength associated with the particle will be .............. $\mathring{A}$.

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The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of the electron becomes $\frac{E}{4}$,then its de Broglie wavelength will be:

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The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is $........$

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If the kinetic energy of a particle in motion is decreased by $36 \%$,the increase in de Broglie wavelength of the particle is (in $\%$)

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The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (kelvin) and mass $m$ is:

Similar Questions

Which of the following graphs correctly represents the de-Broglie wavelength $\lambda$ associated with a fundamental particle having linear momentum $p$?

An $\alpha$-particle moves in a circular path of radius $0.83\, cm$ in the presence of a magnetic field of $0.25\, Wb/m^2$. The de Broglie wavelength associated with the particle will be .............. $\mathring{A}$.

The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of the electron becomes $\frac{E}{4}$,then its de Broglie wavelength will be:

The ratio of the de-Broglie wavelength of an $\alpha$-particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is $........$

If the kinetic energy of a particle in motion is decreased by $36 \%$,the increase in de Broglie wavelength of the particle is (in $\%$)

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