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3 years ago

Conductivity in a metal results from the metal atoms having

Physics

1 answer:

maks197457 [2]3 years ago

5 0

Answer: A) highly mobile electrons in the valence shell

Explanation: conductivity in metals is a result of the movement of electrically charged particles—the electrons. These free electrons also known as valence electrons are free to move, and as a result they can travel through the lattice that forms the physical structure of a metal. The presence of valence electrons determines a metal's conductivity. However, several other factors can affect the conductivity of a metal such as impurities, temperature, magnetic fields etc.

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3 years ago

Why is history open to ongoing and changing interpretations?

saveliy_v [14]

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2 years ago

A book is sitting on a desk. what best describes the normal force acting on the book

Margarita [4]

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3 years ago

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A 5.0 A electric current passes through an aluminum wire of 4.0~\times~10^{-6}~m^2 cross-sectional area. Aluminum has one free e

Serhud [2]

Answer: The electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

Explanation:

Given: Current = 5.0 A

Area = $4.0 \times 10^{-6} m^{2}$

Density = 2.7 $g/cm^{3}$ , Molar mass = 27 g

The electron density is calculated as follows.

$n = \frac{density}{mass per atom}\\= \frac{\rho}{\frac{M}{N_{A}}}\\$

where,

$\rho$ = density

M = molar mass

$N_{A}$ = Avogadro's number

Substitute the values into above formula as follows.

$n = \frac{\rho \times N_{A}}{M}\\= \frac{2.7 g/cm^{3} \times 6.02 \times 10^{23}/mol}{27 g/mol}\\= \frac{16.254 \times 10^{23}}{27} cm^{3}\\= 0.602 \times 10^{23} \times \frac{10^{6} cm^{3}}{1 m^{3}}\\= 6.0 \times 10^{28} m^{-3}$

Thus, we can conclude that the electron number density (the number of electrons per unit volume) in the wire is $6.0 \times 10^{28} m^{-3}$ .

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2 years ago