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 $\left.=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right]$
Correct Option: 1
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)
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