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Question: The mean free path of conduction electrons in copper is about \(4 \times {10^{ - 8}}{\text{m}}\). Fi...

The mean free path of conduction electrons in copper is about 4×10−8m4 \times {10^{ - 8}}{\text{m}}. Find the electric field which can give an average of 2eV2{\text{eV}} energy to a conduction electron in a block of copper.
A) 6×107Vm−16 \times {10^7}{\text{V}}{{\text{m}}^{ - 1}}
B) 7×107Vm−17 \times {10^7}{\text{V}}{{\text{m}}^{ - 1}}
C) 5×107Vm−15 \times {10^7}{\text{V}}{{\text{m}}^{ - 1}}
D) 5×109Vm−15 \times {10^9}{\text{V}}{{\text{m}}^{ - 1}}

Explanation

Solution

The mean free path of the conduction electrons refers to the distance an electron moves under the action of the electric force. The electric force is known to be the product of the charge of the electron and the applied electric field. We can make use of the work-energy theorem which provides the work done by the electric force as the average energy of the electron to find the electric field.

Formulas used:
-The electric force of a charge is given by, F=qEF = qE where qq is the charge and EE is the applied electric field.
-The work done by a force acting on a body is given by, W=FdW = Fd where FF is the applied force and dd is the displacement of the body.

Complete step by step answer.
Step 1: List the parameters given in the question.
The mean free path of the conduction electron is given to be d=4×10−8md = 4 \times {10^{ - 8}}{\text{m}} .
The average energy of an electron in an electric field of strength EE is given to be KE=2eVKE = 2{\text{eV}} .
The charge of an electron is known to be e=1⋅6×10−19Ce = 1 \cdot 6 \times {10^{ - 19}}{\text{C}} .
We have to determine the strength of the electric field.
Step 2: Express the work done by the electric force of an electron.
The electric force of an electron of charge q=eq = e can be expressed as F=eEF = eE ------ (1)
Now the work done by the force to move an electron over a distance dd is given by,
W=FdW = Fd --------- (2)
Substituting equation (1) in (2) we get, W=eEdW = eEd
According to the work-energy theorem, the work done is equal to the kinetic energy or here the average energy of the conducting electron i.e., W=KEW = KE .
So we have the average energy of the electron as KE=eEdKE = eEd
⇒E=KEed\Rightarrow E = \dfrac{{KE}}{{ed}} -------- (3)
Substituting for KE=3⋅2×10−19JKE = 3 \cdot 2 \times {10^{ - 19}}{\text{J}}, e=1⋅6×10−19Ce = 1 \cdot 6 \times {10^{ - 19}}{\text{C}} and d=4×10−8md = 4 \times {10^{ - 8}}{\text{m}} in equation (3) we get, E=3⋅2×10−191⋅6×10−19×4×10−8=5×107Vm−1E = \dfrac{{3 \cdot 2 \times {{10}^{ - 19}}}}{{1 \cdot 6 \times {{10}^{ - 19}} \times 4 \times {{10}^{ - 8}}}} = 5 \times {10^7}{\text{V}}{{\text{m}}^{ - 1}}
Thus we obtain the strength of the electric field as E=5×107Vm−1E = 5 \times {10^7}{\text{V}}{{\text{m}}^{ - 1}} .

So the correct option is C.

Note: While substituting values of different physical quantities in any equation make sure that all the quantities are expressed in their respective S.I units. If this is not so, then the necessary conversion of units must be done. Here the average energy of the conducting electron KE=2eVKE = 2{\text{eV}} is expressed in units of electron-volt. So we converted it into units of Joules as KE=2×1⋅6×10−19=3⋅2×10−19JKE = 2 \times 1 \cdot 6 \times {10^{ - 19}} = 3 \cdot 2 \times {10^{ - 19}}{\text{J}} before substituting in equation (3).