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Page No 510:CONT - Chapter 14 - Semiconductors - Additional Exercise Solutions - Ncert Solutions class 12 - Physics


Question 14.12:

The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenic and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes. Given that ni= 1.5 × 1016 m−3. Is the material n-type or p-type?

Answer:

Number of silicon atoms, N = 5 × 1028 atoms/m3

Number of arsenic atoms, nAs = 5 × 1022 atoms/m3

Number of indium atoms, nIn = 5 × 1020 atoms/m3

Number of thermally-generated electrons, ni = 1.5 × 1016 electrons/m3

Number of electrons, ne = 5 × 1022 − 1.5 × 1016 ≈ 4.99 × 1022

Number of holes = nh

In thermal equilibrium, the concentrations of electrons and holes in a semiconductor are related as:

nenh = ni2

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/258/8059/NS_29-10-08_Sravana_12_Physics_14_19_NRJ_LVN_html_242a1a2e.gif

Therefore, the number of electrons is approximately 4.99 × 1022 and the number of holes is about 4.51 × 109. Since the number of electrons is more than the number of holes, the material is an n-type semiconductor.

Question 14.13:

In an intrinsic semiconductor the energy gap Egis 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration niis given by

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/258/8060/NS_29-10-08_Sravana_12_Physics_14_19_NRJ_LVN_html_55799c4f.gif

where nis a constant.

Answer:

Energy gap of the given intrinsic semiconductor, Eg = 1.2 eV

The temperature dependence of the intrinsic carrier-concentration is written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/258/8060/NS_29-10-08_Sravana_12_Physics_14_19_NRJ_LVN_html_55799c4f.gif

Where,

kB = Boltzmann constant = 8.62 × 10−5 eV/K

T = Temperature

n0 = Constant

Initial temperature, T1 = 300 K

The intrinsic carrier-concentration at this temperature can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/258/8060/NS_29-10-08_Sravana_12_Physics_14_19_NRJ_LVN_html_410d69a8.gif  … (1)

Final temperature, T2 = 600 K

The intrinsic carrier-concentration at this temperature can be written as:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/258/8060/NS_29-10-08_Sravana_12_Physics_14_19_NRJ_LVN_html_2c8ce9a7.gif  … (2)

The ratio between the conductivities at 600 K and at 300 K is equal to the ratio between the respective intrinsic carrier-concentrations at these temperatures.

https://img-nm.mnimgs.com/img/study_content/curr/1/12/16/258/8060/NS_29-10-08_Sravana_12_Physics_14_19_NRJ_LVN_html_m3cd6db40.gif

Therefore, the ratio between the conductivities is 1.09 × 105.