What is the mean time between collisions for electrons in an aluminum wire and in an iron wire?

Answer :

In every collision with associate degree atom, the lepton loses primarily all its forward momentum, thus it's rate once a collision is zero. If the wire there's a current within the wire in response to associate degree obligatory electrical phenomenon, then between collisions, the electrons are accelerated by the electrical field, and a = q*E/m, wherever letter of the alphabet is that the charge on associate degree lepton, m is that the lepton mass, and E is that the force field. If the common time between collisions is T, then the common speed of associate degree lepton simply before it collides is a*T = q*E*T/m, and also the average speed of the electrons throughout the intervals between collisions is (q + q*E*T/m)/2 = q*E*T/(2*m) = v_d. v_d is named the drift rate.

Explanation:

I this in a very wire is just the entire quantity of charge (Q) that passes through a given cross sectional space (A) of the wire per unit time. N physical phenomenon electrons per unit volume(V) within the wire, and these all move with the common drift rate, then this within the wire is given by:

I = dQ/dt = q*N*A*vd = (q^2 * N * T *A *E)/(2*m)

The electric field is just the voltage distinction (V) divided by the length of the wire (L), so:

I = V*(q^2 * N * T * A)/(2*m*L).

For associate degree resistance unit material, we all know phenomenological that V = I*R, thus we will equate R with the reciprocal of the issue that multiplies V within the previous equation:

R = ((2*m)/(q^2 * N * T)) * (L/A)

R = r * L/A

where r = (2*m)/(q^2*N*T) is named the ohmic resistance of the fabric within the wire. ohmic resistance has units of ohm*m.

Rearranging this to resolve for the unit of time between collisions, we have:

T = (2*m)/(q^2 * N * r)

At temperature, the ohmic resistance of Al and two6|metallic element|metal} are 2.65*10^-8 ohm*m and nine.71*10^-8 ohm*m, severally (see source).

The mass of associate degree lepton is nine.109*10^-31 kilo and also the charge on associate degree lepton is one.602*10^-19 C

The only remaining unknown on the r.h.s. of this equation is that the variety of free (conduction) electrons per unit volume. To calculate this, we want to multiply the amount of physical phenomenon atoms per atom (n) by the number density of atoms within the material:

N = (n * Avogadro's variety * density)(atomic mass)

Fe has 2 physical phenomenon electrons per atom, and Al has 3 physical phenomenon electrons per atom

The density of atomic number 26 and Al are seven.87 and 2.7 gm/cm^3, severally, and their atomic lots are fifty five.845 and 26.982 gm/mol, severally.

This gives:

N_Al = 2.58*10^23 electrons/m^3

and

N_Fe = 2.42*10^23 electrons/m^3

Plugging of these values into the on top of equation, one finds (if I did my arithmetic right) that:

T_Al = 1.26*10^-9 s

T_Fe = 3.67*10^-10 s

what question you dont ask a question

explanation:

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over