
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at temperature T (Kelvin) and mass $m$, is:
$A.\dfrac{{2h}}{{\sqrt {mkT} }}$
$B.\dfrac{h}{{\sqrt {mkT} }}$
$C.\dfrac{h}{{\sqrt {3mkT} }}$
$D.\dfrac{{2h}}{{\sqrt {3mkT} }}$
Answer
599.1k+ views
HINT- The concept that matter behaves like a wave was proposed by Louis de Broglie in 1924. It is also referred to as the de Broglie hypothesis. Matter waves are referred to as de Broglie waves.
Complete step-by-step answer:
The de Broglie wavelength is the wavelength, $\lambda $, associated with a massive particle (i.e., a particle with mass, as opposed to a massless particle) and is related to its momentum, p, through the Planck constant, $h$ :
$\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}$ .
De Broglie, in his 1924 PhD thesis, proposed that just as light has both wave- like and particle- like properties, electrons also have wave-like properties. By rearranging the momentum equation, we find a relationship between the wavelength $\lambda $ , associated with an electron and its momentum, p, through the Planck constant, h:
$\lambda = \dfrac{h}{p}$
The relationship is now known to hold for all types of matter: all matter exhibits properties of both particles and waves.
Heavy water (${D_2}O$ ) is a form of water that contains only deuterium rather than the common hydrogen-1 isotope that makes up most of the hydrogen in normal water.
At thermal equilibrium with heavy water, we know that
${E_k} = \dfrac{3}{2}kT$
Now, $p = \sqrt {2m{E_k}} \Rightarrow \lambda = \dfrac{h}{{\sqrt {2m{E_k}} }}$ ($\because \lambda = \dfrac{h}{p}$ )
$\lambda = \dfrac{h}{{\sqrt {2{E_k}m} }} \Rightarrow \lambda = \dfrac{h}{{\sqrt {3mkT} }}$ ($\because {E_k} = \dfrac{3}{2}kT$ )
which gives option $C.$ as the correct option.
NOTE- The physical reality underlying de Broglie waves is a subject of ongoing debate. Some theories treat either the particle or the wave aspect as its fundamental nature, seeking to explain the other as an emergent property. Some, such as the hidden variable theory, treat the wave and the particle as distinct entities. Yet others propose some intermediate entity that is neither quite wave nor quite particle but only appears as such when we measure one or the other property.
Complete step-by-step answer:
The de Broglie wavelength is the wavelength, $\lambda $, associated with a massive particle (i.e., a particle with mass, as opposed to a massless particle) and is related to its momentum, p, through the Planck constant, $h$ :
$\lambda = \dfrac{h}{p} = \dfrac{h}{{mv}}$ .
De Broglie, in his 1924 PhD thesis, proposed that just as light has both wave- like and particle- like properties, electrons also have wave-like properties. By rearranging the momentum equation, we find a relationship between the wavelength $\lambda $ , associated with an electron and its momentum, p, through the Planck constant, h:
$\lambda = \dfrac{h}{p}$
The relationship is now known to hold for all types of matter: all matter exhibits properties of both particles and waves.
Heavy water (${D_2}O$ ) is a form of water that contains only deuterium rather than the common hydrogen-1 isotope that makes up most of the hydrogen in normal water.
At thermal equilibrium with heavy water, we know that
${E_k} = \dfrac{3}{2}kT$
Now, $p = \sqrt {2m{E_k}} \Rightarrow \lambda = \dfrac{h}{{\sqrt {2m{E_k}} }}$ ($\because \lambda = \dfrac{h}{p}$ )
$\lambda = \dfrac{h}{{\sqrt {2{E_k}m} }} \Rightarrow \lambda = \dfrac{h}{{\sqrt {3mkT} }}$ ($\because {E_k} = \dfrac{3}{2}kT$ )
which gives option $C.$ as the correct option.
NOTE- The physical reality underlying de Broglie waves is a subject of ongoing debate. Some theories treat either the particle or the wave aspect as its fundamental nature, seeking to explain the other as an emergent property. Some, such as the hidden variable theory, treat the wave and the particle as distinct entities. Yet others propose some intermediate entity that is neither quite wave nor quite particle but only appears as such when we measure one or the other property.
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