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Question: Given that the mobility of electrons in ge is \(0.4{{m}^{2}}{{V}^{-1}}{{s}^{-1}}\) and electronic ch...

Given that the mobility of electrons in ge is 0.4m2V−1s−10.4{{m}^{2}}{{V}^{-1}}{{s}^{-1}} and electronic charge is 1.6×10−19C1.6\times {{10}^{-19}}C. The number of donor atom semiconductor of conductivity 500mho/m500mho/m is
a)8×1021 b)8×1015 c)5×1021 d)8×1016 \begin{aligned} & a)8\times {{10}^{21}} \\\ & b)8\times {{10}^{15}} \\\ & c)5\times {{10}^{21}} \\\ & d)8\times {{10}^{16}} \\\ \end{aligned}

Explanation

Solution

The conductivity of a metal is the product of the number of donor atoms, mobility of the electrons, and the charge of the electrons. As we know all values except the number of donor atoms, we can easily find it.
Formulas used:
σ=qneμe\sigma =q{{n}_{e}}{{\mu }_{e}}
Where, σ\sigma is conductivity, μe{{\mu }_{e}} is mobility of electrons
Complete step-by-step solution
let us first write down the given terms,
σ=500 ⇒q=1.6×10−19C ⇒μe=0.4m2V−1s−1 \begin{aligned} & \sigma =500 \\\ & \Rightarrow q=1.6\times {{10}^{-19}}C \\\ & \Rightarrow {{\mu }_{e}}=0.4{{m}^{2}}{{V}^{-1}}{{s}^{-1}} \\\ \end{aligned}
Substituting these values in the formula,
σ=qneμe ⇒ne=σqμe ⇒ne=5001.6×10−19×0.4 ⇒ne=7.8×1021 ∼8×1021 \begin{aligned} & \sigma =q{{n}_{e}}{{\mu }_{e}} \\\ & \Rightarrow {{n}_{e}}=\dfrac{\sigma }{q{{\mu }_{e}}} \\\ & \Rightarrow {{n}_{e}}=\dfrac{500}{1.6\times {{10}^{-19}}\times 0.4} \\\ & \Rightarrow {{n}_{e}}=7.8\times {{10}^{21}} \\\ & \sim 8\times {{10}^{21}} \\\ \end{aligned}
Therefore, the correct option is option a.
Additional information:
In physics , the electron mobility tells us how quickly an electron can move through a metal or semiconductor when pulled by an electrical field. There is an analog quantity for holes called the hall mobility. Which term refers to both electron and hole mobility. Electron and hole mobility are special cases for electrical mobility of charged particles during a fluid under an applied field . Conductivity is proportional to the merchandise of mobility and carrier concentration.
There is an easy relation between mobility and electrical conductivity. This formula is valid when conductivity is due entirely to electrons. Hey if a semiconductor has both electrons and holes total conductivity will be the sum of the conductivity of electrons and the conductivity of holes.

Note: This formula which tells us the relation between mobility and electrical conductivity is valid only when the conductivity is entirely due to the electron. In the case of a p-type semiconductor, the conductivity due to holes in the state as there is all hole’s present, the formula is the same.