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== ... (i)
Here, E is the kinetic energy.
E= Substituting value of
E in Eq. (i), we get,
λ= 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.