Question

Find the number of free electron per unit volume of a metallic wire of density $10^{4} \,kg / m ^{3}$ atomic mass number $100$ if the number of free electron per atom is one :-

• A. $6 \times 10^{20}$
• B. $1.6 \times 10^{-19}$
• C. $6.023 \times 10^{28}$
• D. $1.6 \times 10^{19}$

Answer 1

Answer: (C)

Number of atoms in one $m ^{3}$
$=\frac{ M }{ M _{0}} \times N _{ A }=\frac{10^{4}}{100 \times 10^{-3}} N _{ A }$
$=10^{5} \times 6.023 \times 10^{23}=6.023 \times 10^{28}$
Same is the number of free electron.