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

\[(A)0.25A\]
\[(B)0.45A\]
\[(C)0.56A\]
\[(D)0.64A\]

Hint: 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.