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: [$\mathrm{me}=$ 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 $=1.38 \times 10^{-23} \mathrm{JK}^{-1}$]

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Modern Physics

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:
[

me =

mass of electron

= 9 \times 1 0^{- 31} kg

h =

Planck constant

= 6.6 \times 1 0^{- 34} Js

k_{B} =

Boltzmann constant

= 1.38 \times 1 0^{- 23} JK^{- 1}

]

Answer: (B)

The thermal kinetic energy in three dimensions is

E = \frac{3}{2} k_{B} T

, and the de Broglie wavelength is

λ = \frac{h}{2 m _{e} E} = \frac{h}{3 m _{e} k _{B} T}

. Substituting

h = 6.6 \times 1 0^{- 34}

m_{e} = 9 \times 1 0^{- 31}

k_{B} = 1.38 \times 1 0^{- 23}

and

T = 300

gives

λ = \frac{6.6 \times 1 0 ^{- 34}}{3 \times 9 \times 1 0 ^{- 31} \times 1.38 \times 1 0 ^{- 23} \times 300} = 6.26 \times 1 0^{- 9} m

, about

6.26 nm

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Single correct option