Question

Doped silicon has a Hall-coefficient of 3.68 × 10^-4 m³C^-1 and then its carrier concentration value is:

1. 2.0 × 10²² m^-3
2. 2.0 × 10^-22 m-3
3. 0.2 × 10²² m-3
4. 0.2 × 10^-22 m-3

Answer 1

Correct Answer - Option 1 : 2.0 × 10²² m^-3

Concept:

 The relation between the Hall coefficient (Hall constant) and the doping concentration are related as:

\({R_H} = \frac{1}{{ - nq}}\)

\({R_H} = \frac{1}{{ pq}}\)

From these, we can find the electronics and hole concentration.

\({R_H} = \frac{{p\mu _p^2 - n\mu _n^2}}{{q{{\left( {p{\mu _p} + n{\mu _n}} \right)}^2}}}\)

μ = σ R_H

calculation:

Given doping concentration is 3.68 × 10^-4 m³C^-1

Carrier concentration is defined as:

\(n = \frac{1}{{{R_H}q}}\)

\(n = \frac{1}{{3.68 × {{10}^{ - 4}} × 1.6 × {{10}^{ - 19}}}}\)

In ESE prelims no calculator is allowed so, we will do the approximate calculation.

let \(n = \frac{1}{4 × 1.5}\)

\(= \frac{1}{6}\)

= 0.166

We have assumed the more value in the denominator, so decrease the value according to the question, then fraction value increases.

Due to the slight increase value of n maybe

≈1 .7 × 10²²

n = 1.7 × 10²² cm^- 3

No option is matching but we have to choose near value, from the options it is 2.0