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 10^{-34} $Js, $k_b $ = Boltzmann constant = $1.38 \times 10^{-23} JK^{-1} $]

(A) 3.25 nm

(B) 6.26 nm

(C) 2.26 nm

(D) 8.46 nm

*Solution*

de-Broglie wavelength $\lambda = \frac {h}{p} = \frac {h}{\sqrt {2m_e K}}$

For ideal gas in 3-dimensions, $K = \frac {3}{2} k_b T $

$\therefore \lambda = \frac{h}{{\sqrt {3{m_e}{k_b}T} }} = \frac{{6.6 \times {{10}^{ - 34}}}}{{\sqrt {3 \times 9 \times {{10}^{ - 31}} \times 1.38 \times {{10}^{ - 23}} \times 300} }}$

$\lambda \approx \frac{{2.2 \times {{10}^{ - 34}}}}{{3 \times 1.17 \times {{10}^{ - 26}}}} \approx 6.26 \times {10^{ - 9}}m = 6.26nm$

Answer: (B)