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

[m_e = mass of electron = 9 $\times$ 10^$-$31 kg, h = Planck constant = 6.6 $\times$ 6.6 $\times$ 10^$-$34 Js, k_B = Boltzmann constant = 1.38 $\times$ 10^$-$23 JK^$-$1]

Given, Planck's constant, h = 6.6 $\times$ 10^$-$34 Js<br/><br/>Boltzmann constant, k_B = 1.38 $\times$ 10^$-$23 J/K<br/><br/>Mass of an electron, m_e = 9 $\times$ 10^$-$31 kg<br/><br/>Temperature of an ideal gas, T = 300 K<br/><br/>As we know that, de-Broglie wavelength,<br/><br/>$\lambda = {h \over {mv}} = {h \over {\sqrt {2mE} }}$ .... (i)<br/><br/>Here, E is the kinetic energy,<br/><br/>$E = {{3{K_B}T} \over 2}$<br/><br/>Substituting value of E in Eq. (i), we get<br/><br/>$\lambda = {h \over {\sqrt {3m{K_B}T} }}$<br/><br/>Substituting the given values in the above equation, we get<br/><br/>$$\lambda = {{6.6 \times {{10}^{ - 34}}} \over {\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$$<br/><br/>= 6.26 nm<br/><br/>$\therefore$ The corresponding de-Broglie wavelength of an electron approximately at 300 K is 6.26 nm.