Question

If n,e,t and m are representing the electron density, charge, relaxation time and mass of an electron, then resistance of wire with of length l and cross sectional area A is given by:
(A) $\dfrac{{ml}}{{n{e^2}tA}}$
(B) $\dfrac{{2ma}}{{n{e^2}t}}$
(C) $n{e^2}tA$
(D) $\dfrac{{n{e^2}tA}}{{2m}}$

Answer 1

The formula for resistance will have the quantities V and I. We thus need to find a relation between current and drift velocity and relaxation time. We also know how drift velocity is related to the electric field applied and further how electric fields can be expressed in terms of voltage. Manipulating these equations will give us the correct option.

Complete step by step solution
As we know that drift velocity of an electron sitting in an electric field is given by:
${v_d} = \dfrac{{eEt}}{m}$
Where e is the charge of an electron
E is the electric field applied
M is the mass of the electron and
T is the relaxation time; it is defined as the time period between 2 successive collisions between electrons when electric current passes through them.
We also know that the current flowing through an electric conductor is given as:
$I\, = \,neA{v_d}$
Substituting the value of drift velocity in the given equation we get,
$I\, = \,neA(\dfrac{{eEt}}{m})$
As we know that electric field is mathematically equal to the potential applied across unit length
$I\, = \,n{e^2}A(\dfrac{{Vt}}{{lm}})$
$\dfrac{I}{V}\, = \,n{e^2}A(\dfrac{T}{{ml}})$
Now, according to Ohm’s law, potential difference applied across a path is directly proportional to the electric current flowing through it, i.e.
$V \propto I$
$V\, = \,IR$
$\dfrac{V}{I}\, = \,R$
But, $\dfrac{I}{V}\, = \,n{e^2}A(\dfrac{T}{{ml}})$ , Therefore,
$R\, = \,\dfrac{{ml}}{{n{e^2}At}}$

Therefore the option with the correct answer is option A

Note If you already know the relation between conductivity and the quantities n, e, t and m, you can substitute it in $R\, = \,\dfrac{l}{{\sigma A}}$ to get the required solution. This is a shortcut to answer the question.