Question

The de Broglie wavelength of an electron in a metal at ${27}^{\circ }C$ is $(Given{m}_e=9.1\times {10}^{-31}kg,k_B=1.38\times {10}^{-23}J{K}^{-1})$

• A. $6.2\times {10}^{-9}m$
• B. $6.2\times {10}^{-10}m$
• C. $6.2\times {10}^{-8}m$
• D. $6.2\times {10}^{-7}m$

Answer 1

Answer: (A)

$6.2\times {10}^{-9}m$

Here, T = 27 + 273 = 300 K
For an electron in a metal, momentum $p=\sqrt{3m{k}_{B}T}$
de Broglie wavelength of an electron is
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{3m{k}_{B}T}}$
$=\frac{h}{\sqrt{3\times (9.1\times {10}^{-31})\times (1.38\times {10}^{-23})\times 300}}$

$=6.2\times {10}^{-9}m$