First slide

Classification of semi-conductors

Question

A potential difference of 2V is applied between the opposite faces of a Ge crystal plate of area 1 $c m^{2}$ and thickness 0.5 mm. If the concentration of electrons in Ge is $2 \times 10^{19} / m^{3}$ and mobilities of electrons and holes are 0.36 $\frac{m^{2}}{volt - \sec} and 0.14 \frac{m^{2}}{volt - \sec},$ respectively, then the current flowing through the plate will be

Difficult

A

0.25 A
B

0.45 A
C

0.56 A
D

0.64 A

Solution

$\begin{array}{l} σ = ne (μ_{e} + μ_{h}) (∵ n_{e} = n_{h} = n) \\ = 2 \times 10^{19} \times 1 .6 \times 10^{- 19} (0 .36 + 0 .14) = 1 .6 {(Ω - m)}^{- 1} \\ R = ρ \frac{L}{A} = \frac{L}{σA} = \frac{0 .5 \times 10^{- 3}}{1 .6 \times 10^{- 4}} = \frac{25}{8} Ω \\ ∴ i = \frac{V}{R} = \frac{2}{25 / 8} = \frac{16}{25} A = 0 .64 A \end{array}$

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