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:
$\left[\mathrm{m}_{\mathrm{e}}=\right.$ 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 $\left.=1.38 \times 10^{-23} \mathrm{JK}^{-1}\right]$
Select the correct option:
A
6.26 nm
B
8.46 nm
C
2.26 nm
D
3.25 nm
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
De-Broglie wavelength
$$
\lambda=\frac{\eta}{m v}=\frac{h}{\sqrt{2 m E}}
$$
Where E is kinetic energy
$$
\begin{aligned}
& E=\frac{3 \mathrm{kT}}{2} \text { for gas } \\
& \lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 9 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}} \\
& \lambda=6.26 \times 10^{-9} \mathrm{~m}=6.26 \mathrm{~nm}
\end{aligned}
$$
Option (1)
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