
At temperature T, the average kinetic energy of any particle is \[\dfrac{3}{2}RT\]. The de Broglie wavelength follows which of the following order?
(A) visible photon > Thermal electron > thermal neutron
(B) Thermal proton > Visible photon > Thermal neutron
(C) Thermal proton > Thermal electron > Visible photon
(D) visible photon > thermal neutron > Thermal electron
Answer
561.3k+ views
Hint: In order to find the de Broglie wavelength of the particles, we must first know what is a de Broglie equation. De Broglie equation was derived based on the concept that matter has a dual nature, i.e. matter can behave as a particle as well as a wave. He derived the following equation.
\[\lambda = \dfrac{h}{{mv}}\]
Complete Solution :
In this problem, it is given that at temperature T, the average kinetic energy of the particles is given as
\[K.E = \dfrac{3}{2}RT\]
De Broglie equation was derived based on the concept that matter has a dual nature, i.e. matter can behave as a wave as well as a particle. The de Broglie equation is given as
\[\lambda = \dfrac{h}{{mv}}\]
‘\[mv\]’ can be obtained from the kinetic energy of the particles.
\[K.E = \dfrac{1}{2}m{v^2}\]
\[\lambda = \dfrac{h}{{\sqrt {2mK.E} }}\]
\[\lambda = \dfrac{h}{{\sqrt {2m} \frac{3}{2}RT}}\]
\[\lambda \propto \dfrac{1}{m}\]
Therefore, wavelength is inversely proportional to mass of the particles.
We know that the mass of the neutron will be greater than the mass of the electron. Also the mass of the photon being zero can be negligible.
Therefore, mass of the particles is arranged in the following order.
\[{m_{neutron}} > {m_{electron}} > {m_{photon}}\]
As the wavelength is inversely proportional to the mass of the particles. The above given order is reversed.
The de Broglie wavelength will be given in the following order
visible photon > Thermal electron > thermal neutron.
So, the correct answer is “Option A”.
Note: The de Broglie waves can have various applications such as:
- The wave properties are only observed in very small objects. The de Broglie wavelength is produced only when the electron is used as a source.
- The wavelength used will limit the size of the particle that can be seen in the microscope.
- By using the electron microscope, we can study the fine details of the smallest particles.
\[\lambda = \dfrac{h}{{mv}}\]
Complete Solution :
In this problem, it is given that at temperature T, the average kinetic energy of the particles is given as
\[K.E = \dfrac{3}{2}RT\]
De Broglie equation was derived based on the concept that matter has a dual nature, i.e. matter can behave as a wave as well as a particle. The de Broglie equation is given as
\[\lambda = \dfrac{h}{{mv}}\]
‘\[mv\]’ can be obtained from the kinetic energy of the particles.
\[K.E = \dfrac{1}{2}m{v^2}\]
\[\lambda = \dfrac{h}{{\sqrt {2mK.E} }}\]
\[\lambda = \dfrac{h}{{\sqrt {2m} \frac{3}{2}RT}}\]
\[\lambda \propto \dfrac{1}{m}\]
Therefore, wavelength is inversely proportional to mass of the particles.
We know that the mass of the neutron will be greater than the mass of the electron. Also the mass of the photon being zero can be negligible.
Therefore, mass of the particles is arranged in the following order.
\[{m_{neutron}} > {m_{electron}} > {m_{photon}}\]
As the wavelength is inversely proportional to the mass of the particles. The above given order is reversed.
The de Broglie wavelength will be given in the following order
visible photon > Thermal electron > thermal neutron.
So, the correct answer is “Option A”.
Note: The de Broglie waves can have various applications such as:
- The wave properties are only observed in very small objects. The de Broglie wavelength is produced only when the electron is used as a source.
- The wavelength used will limit the size of the particle that can be seen in the microscope.
- By using the electron microscope, we can study the fine details of the smallest particles.
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