JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given, Planck's constant, h=6.6×10−34‌Js
Boltzmann constant, kB=1.38×10−23J∕K
Mass of an electron, me=9×10−31‌kg
Temperature of an ideal gas, T=300K
As we know that, de-Broglie wavelength,
λ=

h

mv

=

h

√2‌mE

... (i)
Here, E is the kinetic energy.
E=

3KBT

2

Substituting value of E in Eq. (i), we get,
λ=

h

√3mKBT

Substituting the given values in the above equation, we get
λ=

6.6×10−34

√3×9×10−31×138×10−23×300

= 6.26 nm
∴ The corresponding de-Broglie wavelength of an electronapproximately at 300 K is 6.26 nm.