Question

Calculate the number of states per cubic meter of sodium in 3s band. The density of sodium is \[1013{\text{ }}kg{m^{ - 3}}\]. How many of them are empty?

Answer 1

Hint: The given problem is from semiconductor electronics. To solve this first we calculate the total number of atoms. Then after considering electronic configuration, we will calculate the number of states and number of empty states.

Complete step by step answer:
According to given problem the data is given as
Density of sodium, d= \[1013{\text{ }}kg{m^{ - 3}}\]
Volume, V= $1{m^3}$
Mass can be defined as the product of density and volume, So here we will calculate the mass of sodium as,
$m = d \times V = 1013 \times 1 = 1013kg$
Molecular mass of sodium, M=23

As we know that 23 gram sodium contains 6 atoms, so the number of atoms in 1013 kg sodium will be-
$
  N = \dfrac{{1013 \times {{10}^3} \times 6 \times {{10}^{23}}}}{{23}} \\
   \Rightarrow N = 264.26 \times {10^{26}} \\
 $
(A) As the number of maximum possible electrons that can occupy the 3s band is 2, the total number of seats in the 3s band will be
$
  N = 2 \times 264 \times {10^{26}} \\
   \Rightarrow N = 5.3 \times {10^{28}} \\
 $
(B) As the atomic number of sodium is 11, its electronic configuration is $1{s^2},2{s^2},2{p^6},3{s^1}$.

According to electron filling in different bands, s band requires 2 electrons in itself. But This implies that the 3s band is half filled in case of sodium, so the total number of unoccupied states is half of the total number of seats in the 3s band.
Number of empty states=$\dfrac{{5.3 \times {{10}^{28}}}}{2} = 2.65 \times {10^{28}}$
These are the numbers of states that are empty in sodium atoms.
This is the required solution for a given problem.

Note: The electron configuration of an element describes how electrons are distributed in its atomic orbitals. Electron configurations are useful for determining the valency of an element, interpreting atomic spectra and predicting the properties of a group of elements.