Note that thermally generated electrons (ni ~10¹⁶m^–3) are negligibly small as compared to those produced by doping.
Therefore, ne \(\approx\) ND
Since n_e n_h = \(n_{i}^{2}\) , The number of holes
n_h = (2.25 x 10³² ) / (5 x10²²)
= ~ 4.5 x 10⁹ m^–3

Doping is a process of deliberate addition of a desirable impurity in a pure semiconductor to modify its properties in a controlled manner.

Saturation/Short nature of nuclear force

\(R=R_0 A^{1 / 3}, R_1 / R_2=A_1^{1 / 3} / A_2^{1 / 3}=1,{ }^{1 / 3} / 27_{1 / 3}=1 / 3\)

Radius of the nth Bohr orbit \(r=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m Z e^{2}}\)
Again, \(r \propto \frac{1}{Z}\)
\(\therefore\) \(\frac{r_{\mathrm{He}^{+}}}{r_{\mathrm{H}}}=\frac{Z_{\mathrm{H}}}{Z_{\mathrm{He}^{+}}}\)
For hydrogen, Z = 1 and for helium, Z = 2
\(\therefore\) \(\frac{r_{\mathrm{He}^{+}}}{r_{\mathrm{H}}}=\frac{1}{2}\)
\(\Rightarrow \quad r_{\mathrm{He}^{+}}=\frac{1}{2} r_{\mathrm{H}}=\frac{0.53}{2}=0.265 \) \(\overset{o}{A}\)
For radius of third orbit, i.e. for n = 3
r³ = (3)² x 0.265 \(\overset{o}{A}\)
= 9 x 0.265\(\overset{o}{A}\)
= 2.38\(\overset{o}{A}\)

\(0.95\times { 10 }^{ -13 }m\)

\(3646.8\mathring { A } \)
\(Take \ { n }_{ 1 }=2 \ and \ { n }_{ 2 }=\infty \)

Helium : 8.79 MeV / nucleon.

Nuclear Fission The phenomenon of splitting of heavy nuclei (mass number> 120) into smaller nuclei of nearly equal masses is known as nuclear fission.
In nuclear fission, the sum of the masses  Y of the product is less than the sum of masses of the reactants. This difference of mass gets converted into energy E = me! and hence sample amount of energy is released in a nuclear fission.
e.g. \(_{ 235 }^{ 92 }{ U }+_{ 0 }^{ 1 }{ n }\rightarrow _{ 56 }^{ 141 }{ Ba }+_{ 36 }^{ 92 }{ Kr }+6_{ 0 }^{1 }{ n }+Q\)
Masses of reactant
= 235.0439 amu + 1.0087 amu
= 236.0526 amu
Masses of product
= 140.9139 + 91.8973 + 3.0261
= 235.8373 amu
Mass defect = 236.0526 - 235.8373
= 0.2153 amu
\(\because\)  1amu \(\equiv\) 931 MeV
\(\Rightarrow\) Energy released = 0.2153 x 931 = 200 MeV nearly
Thus, energy is liberated in nuclear fission \(_{92}^{235}{U}.\)

The donor energy level E_D is just below the bottom of the conduction band. At room temperature this small energy gap is easily converted by the thermally excited electrons. The conduction band has more electrons as they have been contributed both by thermal excitation and donor impurities. Whereas the acceptor energy level  E_A lies slightly above the top of the valence band. At room temperature, many electrons of the valence band get excited to these acceptor energy levels, leaving behind equal number of holes in the valence band. These holes can conduct current. Thus, the valence band has more holes than the electrons in the conduction band.

equal

\(we \ know, \ 1 \ amu=1.66\times { 10 }^{ -27 }kg\)
\(From \ E={ mc }^{ 2 }=(1.66\times { 10 }^{ -27 }){ \left( 3\times { 10 }^{ 8 } \right) }^{ 2 }\)
\( =1.494\times { 10 }^{ -10 }J\)
\(E=\frac { 1.494\times { 10 }^{ -10 } }{ 1.6\times { 10 }^{ -13 } } MeV=933.75MeV\)
\(For \ oxygen_{ 8 }{ { O }^{ 16 } }, \ mass \ defect\)
\(=8{ m }_{ p }+8{ m }_{ n }-{ M }_{ oxy }\)
\(=8\times 1.00727+8\times 1.00866-15.99053\)
\( =0.13691 \ amu\)
\(=0.13691\times 933.75MeV/{ c }^{ 2 }\)
\( =127.8MeV/{ c }^{ 2 }\)

Here, energy released/fission
= 200MeV
\(=200\times 1.6\times { 10 }^{ -13 }J=3.2\times { 10 }^{ -11 }J\)
\( Total \ energy \ required/sec=1 \ kW=1000w\)
\( =1000J/s.\)
\( Number\ of \ fissions/sec\)
\( =\frac { Energy \ reqd.\ per \ sec. }{ energy \ released/fission }\)
\(=\frac { 1000 }{ 3.2\times { 10 }^{ -11 } } =3.125\times { 10 }^{ 13 }s\)

For each atom doped with arsenic, one free electron is received. Similarly, for each atom doped of indium, a vacancy is created. So, number of free electrons introduced by pentavalent impurity is
\(N_{\mathrm{As}}=5 \times 10^{22} \mathrm{~m}^{-3}\)
The number of holes introduced by trivalent impurity added is
\(N_{\mathrm{I}}=5 \times 10^{20} \mathrm{~m}^{-3}\)
So, net number of electrons added is
n_e = N_As - N₁
\(\begin{aligned} =5 \times 10^{22}-5 \times 10^{20} \end{aligned}\)
\(\begin{aligned} =4.95 \times 10^{22} \mathrm{~m}^{-3} \end{aligned}\)
 We know that, \(n_e n_h=n_i^2\)
So,  \(n_h=\frac{n_i^2}{n_e}=\frac{\left(1.5 \times 10^{16}\right)^2}{4.95 \times 10^{22}}\)
\(=4.54 \times 10^9 \mathrm{~m}^{-3}\)
As, n_e > n_h (number of holes). So, the material is n-type semiconductor.