Estimate the average drift speed of conduction electrons in . . .

Question : Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area 1.0×10^–7 m² carrying a current of 1.5 A. Assume that each copper atom contributes roughly one conduction electron. The density of copper is 9.0×10³ kg/m³, and its atomic mass is 63.5 u.

Doubt by Saatvik

Solution : 

A= 1.0×10^–7 m²
I=1.5 A

ρ=9.0×10³ kg/m³

Atomic Mass of Copper (M) 
= 63.5 u.
=63.5 g/mol
=63.5×10^-3 kg/mol

Density of Copper (ρ)=9.0×10³ kg/m³
It measn that 1m³ of copper has mass of 9×10³ Kg 

So consider it as given mass (m) = 9×10³ kg
No. of moles 
= given mass (m) / molar mass (M)

= 9×10³ / 63.5×10^-3
= (9/63.5)×10⁶

1 Mole of copper has 6.022×10²³ atoms 

So 
(9/63.5)×10⁶ moles of copper has 

=(9/63.5)×10⁶×6.022×10²³ atoms
=[(9×6.022)/63.5] ×10²⁹ atoms
It is given that each copper atom provides one electrons, so total number of electrons in 1m³ of copper would be 
=[(9×6.022)/63.5] ×10²⁹ electrons/m³

So, number of electrons per unit volume would be (n) = [(9×6.022)/63.5] ×10²⁹ electrons/m³

In short, 
We have used the following formula 
n=ρVN_A/M

Now, 
We know,
I=neV_dA
I/neA = V_d
V_d=I/neA
V_d=1.5/{[(9×6.022)/63.5]}×10²⁹×1.6×10^-19×1.0×10^–7]}
V_d=1.5×63.5/{[9×6.022×1.6]}×10^-29×10⁺¹⁹×10⁺¹⁷
V_d=[(1.5×63.5)/(9×6.022×1.6]×10^-29+19+17
V_d=[95.25/86.7168]×10^-3

V_d=[1.098]×10^-3
V_d≈1.1×10^-3 m/s or 1.1 mm/s.

Hence, the average drift speed of conduction electrons in a copper wire is approximately 1.1 mm/s.