Question

A potential difference $2V$ is applied between the opposite faces of a Ge crystal plate of area 1 cm and thickness 0.5 mm. If the concentration of electron in Ge is $2 \times {10^{19}}/m$ and mobility of electrons and holes are $0.36\dfrac{{{m^2}}}{{volt - \sec }}$ and $0.14\dfrac{{{m^2}}}{{volt - \sec }}$ respectively, then the current flowing through the plate will be

$(A)0.25A$
$(B)0.45A$
$(C)0.56A$
$(D)0.64A$

Answer 1

Use ohm’s law to find the current flowing through the plate. Ohm’s law states that the current passing through a conductor is directly proportional to the potential difference between the two points. The constant is known as Resistance and it is the property of a material to resist the flow of electric current through a material.

Formula used:
$V = IR$
$V$ is the potential difference between the two ends of the wire,$I$ is current, and $R$ is the resistance of the wire.
$R = \dfrac{{\rho l}}{A}$
$\rho =$ the resistivity of the material, $L\; =$ length of the wire, and $A\; =$ cross-section area of the wire.
Conductivity ($\sigma$ ) is the reciprocal of resistivity,
$\rho = \dfrac{1}{\sigma }$

Complete step-by-step solution:
Ohm’s law can be written as,
$V = IR$
Where, $V$ is the potential difference between the two ends of the wire,$I$ is current, and $R$ is the resistance of the wire.
The resistance of a wire can be given by the formula-
$R = \dfrac{{\rho l}}{A}$
Here, $\rho =$ the resistivity of the material, $L\; =$ length of the wire, and $A\; =$ cross-section area of the wire.
Conductivity ($\sigma$ ) is the reciprocal of resistivity,
$\rho = \dfrac{1}{\sigma }$
this value of resistivity can be substituted in the formula of resistance,
$R = \dfrac{L}{{\sigma A}}$
From the given data,
${\mu _e} = 0.36\dfrac{{{m^2}}}{{volt - \sec }}$, ${\mu _h} = 0.14\dfrac{{{m^2}}}{{volt - \sec }}$, $n = 2 \times {10^{19}}/m$, $l = 0.5mm$, and $V = 2V$.
$\sigma = ne\left( {{\mu _e} + {\mu _h}} \right)$
$\Rightarrow \sigma = 1.6{(\Omega m)^{ - 1}}$
$\Rightarrow \sigma = 2 \times {10^{19}} \times 1.6 \times {10^{ - 19}}(0.36 + 0.14)$
$\therefore R = \dfrac{L}{{\sigma A}}$
$\Rightarrow R = \dfrac{{0.5 \times {{10}^{ - 3}}}}{{1.6 \times {{10}^{ - 4}}}}$
$\Rightarrow R = \dfrac{{25}}{8}\Omega$
Now the current, $I = \dfrac{V}{R} = \dfrac{{2 \times 8}}{{25}} = \dfrac{{16}}{{25}}A = 0.64A$
Hence, option D is correct.

Note: The quantities used in Ohm’s law are all scalar. To convert them into vector form, the resistance is written in the form of resistivity, length, and area of cross-section. The quantities are then rearranged to convert the potential difference into the potential gradient and the current into current density, which is a vector quantity.