# The temperature of an ideal gas in 3 -dimensions is 300K.

Question:

The temperature of an ideal gas in 3 -dimensions is $300 \mathrm{~K}$. The corresponding de-Broglie wavelength of the electron approximately at $300 \mathrm{~K}$, is:

${\left[\mathrm{m}_{\mathrm{e}}=\right.$ mass of electron $=9 \times 10^{-31} \mathrm{~kg}}$

$\mathrm{h}=$ Planck constant $=6.6 \times 10^{-34} \mathrm{Js}$

$\mathrm{k}_{\mathrm{B}}=$ Boltzmann constant $\left.=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right]$

1. $6.26 \mathrm{~nm}$

2. $8.46 \mathrm{~nm}$

3. $2.26 \mathrm{~nm}$

4. $3.25 \mathrm{~nm}$

Correct Option: 1

Solution:

De-Broglie wavelength

$\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$

Where $\mathrm{E}$ is kinetic energy

$\mathrm{E}=\frac{3 \mathrm{kT}}{2}$ for gas

$\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}$

$\lambda=6.26 \times 10^{-9} \mathrm{~m}=6.26 \mathrm{~nm}$

Option (1)