Question

Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10–10 m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distintguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]

Answer 1

Given: The mean separation between two electrons in a metal is 2× 10 −10  m and the temperature is 27 °C

de Broglie wavelength of an electron is given as,

λ= h 3mkT (1)

Where, Planck’s constant is h, the mass of electron is m, Boltzmann Constant is k, the temperature is T and the wavelength is λ.

By substituting the given values in the above formula, we get

λ= 6.6× 10 −34 3×9.1× 10 −31 ×1.38× 10 −23 ×300 = 6.6× 10 −34 11302.2× 10 −54 = 6.6× 10 −34 106.311× 10 −27 =6.2× 10 −9  m

Thus, the de Broglie wavelength of electron is 6.2× 10 −9  m which is much greater than the given inter-electron separation.