Find the number of free electron per unit volume of a metallic wire of density 104 kg / m 3 atomic mass number 100 if the number of free electron per atom is one :-

Question

Find the number of free electron per unit volume of a metallic wire of density 104 kg / m 3 atomic mass number 100 if the number of free electron per atom is one :- (A) 6 × 1020 (B) 1.6 × 10-19 (C) 6.023 × 1028 (D) 1.6 × 1019

Tardigrade Admin · Accepted Answer

Number of atoms in one m 3 =( M / M 0) × N A =(104/100 × 10-3) N A =105 × 6.023 × 1023=6.023 × 1028 Same is the number of free electron.

Q. Find the number of free electron per unit volume of a metallic wire of density $1 0^{4} k g / m^{3}$ atomic mass number $100$ if the number of free electron per atom is one :-

$6 \times 1 0^{20}$

$1.6 \times 1 0^{- 19}$

$6.023 \times 1 0^{28}$

$1.6 \times 1 0^{19}$

Number of atoms in one $m^{3}$
$= \frac{M}{M _{0}} \times N_{A} = \frac{1 0 ^{4}}{100 \times 1 0 ^{- 3}} N_{A}$
$= 1 0^{5} \times 6.023 \times 1 0^{23} = 6.023 \times 1 0^{28}$
Same is the number of free electron.

Q. Find the number of free electron per unit volume of a metallic wire of density 104kg/m3 atomic mass number 100 if the number of free electron per atom is one :-

6×1020

1.6×10−19

6.023×1028

1.6×1019