Question

If the ratio of the concentration of electrons to that of holes in a semiconductor is \[\dfrac{7}{5}\] and the ratio of the currents is \[\dfrac{7}{4}\], then the ratio of drift velocities is
(A) \[\dfrac{5}{8}\]
(B) \[\dfrac{4}{5}\]
(C) \[\dfrac{5}{4}\]
(D) \[\dfrac{4}{7}\]

Answer 1

Hint: Drift velocity is directly proportional to the current and inversely proportional to the carrier density. We need to divide the electron drift velocity by the hole drift velocity.
Formula used: In this solution we will be using the following formulae;
\[I = nqvA\] where \[I\] is the current flowing through a conductor or semiconductor due to a particular carrier, \[n\] is the carrier density, \[q\] is the charge of the carrier, \[v\] is the drift velocity and \[A\] is the cross sectional area of the material

Complete Step-by-Step solution:
To solve the above, generally, we know the current flowing through the semiconductor due to a particular carrier whether holes or electrons would be given by
\[I = nqvA\]where \[I\] is the current flowing through a conductor or semiconductor due to a particular carrier, \[n\] is the carrier density or concentration, \[q\] is the charge of the carrier, \[v\] is the drift velocity and \[A\] is the cross sectional area of the material
Hence, by rearranging to make the drift velocity subject of the formula, we have
\[v = \dfrac{I}{{nqA}}\]
Hence, for electron drift velocity we have,
\[{v_e} = \dfrac{{{I_e}}}{{{n_e}eA}}\]
And for holes, we have
\[{v_h} = \dfrac{{{I_h}}}{{{n_h}eA}}\]
Hence, dividing the electron drift velocity by that of the holes, we have
\[\dfrac{{{v_e}}}{{{v_h}}} = \dfrac{{{I_e}}}{{{n_e}eA}} \div \dfrac{{{I_h}}}{{{n_h}eA}}\]
Hence,
\[\dfrac{{{v_e}}}{{{v_h}}} = \dfrac{{{I_e}}}{{{n_e}eA}} \times \dfrac{{{n_h}eA}}{{{I_h}}}\]
\[ \Rightarrow \dfrac{{{v_e}}}{{{v_h}}} = \dfrac{{{I_e}}}{{{I_h}}} \times \dfrac{{{n_h}}}{{{n_e}}}\]
This can be written as
\[\dfrac{{{v_e}}}{{{v_h}}} = \dfrac{{{I_e}}}{{{I_h}}} \div \dfrac{{{n_e}}}{{{n_h}}}\]
Hence, inserting the given ratios, we have
\[\dfrac{{{v_e}}}{{{v_h}}} = \dfrac{7}{4} \div \dfrac{7}{5}\]
\[ \Rightarrow \dfrac{{{v_e}}}{{{v_h}}} = \dfrac{7}{4} \times \dfrac{5}{7} = \dfrac{5}{4}\]

Hence, the correct option is C

Note: Alternatively, without the knowledge of the entire formula, but noting that the drift velocity is proportional to current but inversely proportional to the charge density, we can have that
\[v \propto \dfrac{I}{n} = k\dfrac{I}{n}\]
Hence, for electrons,
\[{v_e} = k\dfrac{{{I_e}}}{{{n_e}}}\] and for holes, we have
\[{v_h} = k\dfrac{{{I_h}}}{{{n_h}}}\]
When dividing again, we have
\[\dfrac{{{v_e}}}{{{v_h}}} = k\dfrac{{{I_e}}}{{{n_e}}} \div k\dfrac{{{I_h}}}{{{n_h}}}\]
Which by simplification will lead us to the same relation
\[\dfrac{{{v_e}}}{{{v_h}}} = \dfrac{{{I_e}}}{{{I_h}}} \div \dfrac{{{n_e}}}{{{n_h}}}\] as given above in solution step.