Discussion
Find the concentration of holes (pp) and electronics (np) in a type germanium at 300 K, if the conductivity is 100 Ω cm−1, hole mobility (μp) = 1800 cm2/V sec, intrinsic charges carrier concentration (ni) = 2.25 × 1013/cm3, charge on electron e = 1.6 × 10−19 C. Assume that the conductivity due to electron is negligible as compared to that due to holes.
Answer: B
The keyed values are \(p_{p} = 3.41 \times 10^{17}/cm^{3}\), \(n_{p} = 1.83 \times 10^{9}/cm^{3}\).
With the electron contribution neglected, the conductivity is \(\sigma = p\,e\,\mu_{p}\), and the minority electrons follow from the mass-action law \(np = n_{i}^{2}\):
\[ \begin{aligned} p_{p} &= \frac{\sigma}{e\mu_{p}} \\ &= \frac{100}{1.6 \times 10^{-19} \times 1800} \\ &\approx 3.5 \times 10^{17}\ cm^{-3} \\ n_{p} &= \frac{n_{i}^{2}}{p_{p}} \sim 10^{9}\ cm^{-3} \end{aligned} \]
The keyed option is the only one with a hole density near \(3.4 \times 10^{17}\) and an electron density many orders smaller. The printed figures do not reproduce exactly from the given data: \(n_{p} = 1.83 \times 10^{9}\) corresponds to \(n_{i} = 2.5 \times 10^{13}\), the usual value for germanium, rather than the stated \(2.25 \times 10^{13}\), which gives about \(1.5 \times 10^{9}\).
The option with equal hole and electron densities ignores the mass-action law, and hole densities of \(10^{5}\) or \(10^{19}\) do not give a conductivity of 100 with this mobility.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development