Find the typical de-Broglie wavelength of an electron in a metal at $27^{o} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{- 10} m$

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Solution

It cap be obtained that de-Broglie wavelength of the electron,
$λ = 62.3 \times 10^{- 10} m = 62.3 A$
Mean separation between two electrons in the metal,
$d = 2 \circ A$
$∴ \frac{λ}{d} = \frac{62.3}{2} = 31$
i.e de-Broglie wavelength of the electron is much larger than the separation between two electrons in the metal.

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Find the typical de-Broglie wavelength of an electron in a metal at 27oC and compare it with the mean separation between two electrons in a metal which is given to be about 2×10−10m

It cap be obtained that de-Broglie wavelength of the electron,λ=62.3×10−10m=62.3AMean separation between two electrons in the metal,d=2∘A∴λd=62.32=31i.e de-Broglie wavelength of the electron is much larger than the separation between two electrons in the metal.

Find the typical de-Broglie wavelength of an electron in a metal at $27^{o} C$ and compare it with the mean separation between two electrons in a metal which is given to be about $2 \times 10^{- 10} m$