Question

De-Broglie wavelength of an atom at absolute temperature $T\;{\rm{K}}$ will be
(A) $\dfrac{h}{{\sqrt {3mKT} }}$
(B) $\dfrac{h}{{mKT}}$
(C) $\dfrac{{\sqrt {2mKT} }}{h}$
(D) $\sqrt {2mKT}$

Answer 1

De-Broglie wavelength provides information about wave nature of a matter. It gives the relation between the wavelength and the momentum of the particle. According to De-Broglie the wavelength of the particle has an inverse relation with the momentum of the particle.

Complete step by step answer:
The given absolute temperature of the atom is $T\;{\rm{K}}$ .
The relation between the wavelength and the momentum of the particle given by De-Broglie is given as follows,
$\lambda = \dfrac{h}{P}$
Here, h is the Planck’s constant, $\lambda$ is the wavelength and $P$ is the momentum.
It is known that the relation between the kinetic energy of the particle and the absolute temperature is,
$E = \dfrac{3}{2}KT...............{\rm{(1)}}$
Here, the kinetic energy of the particle is denoted by$E$, $K$ is the Boltzmann constant and $T$ represents the absolute temperature.
The relation between the kinetic energy and the momentum of the particle is given as follows,
$E = \dfrac{{{P^2}}}{{2m}}..................{\rm{(2)}}$
Here, $P$ is the momentum of the particle and$m$ is the mass of the particle.
Substitute the value of kinetic energy in equation (2) from equation (1).
$\dfrac{3}{2}KT = \dfrac{{{P^2}}}{{2m}}$
Now we will get the value of momentum from the above relation.
${P^2} = 3mKT$
We will take root both side,
$P = \sqrt {3mKT}$
Now, using the De-Broglie relation we will calculate the value of the wavelength of the particle.
$\lambda = \dfrac{h}{{\sqrt {3mKT} }}$
Therefore, the value of the wavelength of an atom at absolute temperature is $\dfrac{h}{{\sqrt {3mKT} }}$ and the correct option is ${\rm{option}}\;{\rm{(A)}}$ .

Note: The matter shows wave- particle duality relation that means at one moment the matter shows the wave behaviour and at another moment it acts like a particle. De-Broglie used the wave behaviour of the matter and gave the relations between the momentum and the wavelength of the particle.