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

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.