Question

Atomic masses of He and Ne are 4 and 20 amu respectively. The value of the de-Broglie wavelength of He gas $- 73^\circ C$ is “M” times that of the de Broglie wavelength of Ne at $727^\circ C$.

Answer 1

The wavelength of the substance which relates to the momentum of the substance and the mass of the substance is said as de Broglie wavelength. The thermal de Broglie wavelength is considered the average de Broglie wavelength is shown by a gas particle in an ideal gas system at a specific temperature.

Complete step by step answer:
Given
The atomic mass of helium is 4 amu.
The atomic mass of neon is 20 amu.
The temperature of He gas is $- 73^\circ C$= 200 K
The temperature of Ne gas is $727^\circ C$= 1000 K.
The de Broglie wavelength of a gas molecule is given by the formula as shown below.
$\lambda = \dfrac{h}{{\sqrt {3mkT} }}$
Where,
$\lambda$ is the wavelength.
h is the planck's constant.
k is the Boltzmann constant.
T is the temperature
It is given that the value of the de-Broglie wavelength of He gas $- 73^\circ C$ is “M” times that of the de Broglie wavelength of Ne at $727^\circ C$.
It is shown as
${\lambda _{He}} = M{\lambda _{Ne}}$
$\Rightarrow \dfrac{h}{{\sqrt {3 \times 4 \times k \times 200} }} = M\dfrac{h}{{\sqrt {3 \times 20 \times k \times 1000} }}$
$\Rightarrow M = \sqrt {25}$
$\Rightarrow M = 5$

Hence, the de Broglie wavelength of helium is 5 times the de Broglie wavelength of neon.

Note: The de Broglie wavelength of the particle is inversely proportional to the force. The matter has both wave nature and particle nature, the wave nature is observed for very small substances. The double-slit interference pattern having the de Broglie wavelength is formed by using electrons as its main source.