Total energy of the electron in hydrogen atom is –13.6 eV = –13.6 × 1.6 × 10^–19 J = –2.2 ×10^–18 J. Thus from Equation we have
\(-\frac{e^{2}}{8 \pi \varepsilon_{0} r}=-2.2 \times 10^{-18} \mathrm{~J}\)
This gives the orbital radius
\(r=-\frac{e^{2}}{8 \pi \varepsilon_{0} E}=-\frac{\left(9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2}}{(2)\left(-2.2 \times 10^{-18} \mathrm{~J}\right)}\)
= 5.3 × 10^–11 m
The velocity of the revolving electron can be computed from Eq.uation with m = 9.1 × 10^–31 kg,
\(v=\frac{e}{\sqrt{4 \pi \varepsilon_{0} m r}}=2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Repulsive force of magnitude 6 x 10⁻³ N
Charge on the first sphere, q₁ = 2 x 10⁻⁷ C
Charge on the second sphere, q₂ = 3 x 10⁻⁷ C
Distance between the spheres, r = 30 cm = 0.3 m
Electrostatic force between the spheres is given by the relation,
\(F=\frac{q_{1} q_{2}}{4 \pi \in_{0} r^{2}}\)
Where, ∈₀ = Permittivity of free space
\(\frac{1}{4 \pi \in_{0}}=9 \times 10^{9} N m^{2} C^{-2}\)
\(F=\frac{9 \times 10^{9} \times 2 \times 10^{-7} \times 3 \times 10^{-7}}{(0.3)^{2}}=6 \times 10^{-3} N\)
Hence, force between the two small charged spheres is 6 x 10⁻³ N. The charges are of same nature. Hence, force between them will be repulsive.

The given diagram shown below

The circuit above can be redrawn as follows

As the p-section is connected to negative terminal of the battery, the diode shown is reverse biased.
(ii) During the first half of input cycle, the upper end of the coil is at positive potential and lower end at negative potential. The function diode DI is forward biased and D₂ in reverse biased. Current flows in output load in the
direction shown in figure. During the second half of input cycle, D₂ is forward biased. In this way, current flows in the load in the single direction as shown in figure.

Given figure shows a schematic diagram of Geiger -Marsden experiment.


The size of the nucleus can be obtained by finding impact parameter b using trajectories of n-particle. The impact parameter is the perpendicular distance of the initial velocity vector of o-particle from the central line of nucleus when it is far away from the atom.
Rutherford calculated impact parameter as
\(b=\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } .\frac { Z{ e }^{ 2 }\cot { \left( { \theta }/{ 2 } \right) } }{ E } \)
where, 
E = KE of (\(\alpha\) -particle)
\(\theta\) = scattering angle
Z = atomic number of atom
The size of the nucleus is smaller than the impact parameter.
The idea of the size of the nucleus can also be obtained by finding the distance of closest approach.

(a) metals, (b) insulators and (c) semiconductors
Two distinguishing features:
(i) In conductors, the valence band and conduction band tend to overlap (or nearly overlap) while in insulators they are separated by a large energy gap and in semiconductors they are separated by a smaJ1'ep.ergygap.  
(ii) The conduction band, of a conductor, has a large number of electrons available for electrical conduction. However the conduction band of insulators is almost empty while that of the semiconductor has only a (very) small number of such electrons available for electrical conduction.

Let us consider two uniformly charge parallel sheets carrying surface charge densities + \(\sigma \) and -\(\sigma \) respectively and are separated by a small distance from each other.
   
By Gauss' law, it can be proved that, electric field intensity due to a uniformly charged infinite plane sheet as nearby is given by
E = \(\frac { \sigma }{ 2{ \varepsilon }_{ 0 } } \)   ....(i) 
The electric field is directed normally outward from the plane sheet, if nature of charge on sheet is positive and normally inward, if charge is of negative nature. Let \(\widehat { r } \) represents unit vector directed from positive plate to negative plate.
Now, electric field intensity (EFI) at any point P between the two plates is given by
(i)   E₁ = + \(\frac { \sigma }{ 2{ \varepsilon }_{ 0 } } \widehat { r } \)   [due to positive plate]
(ii)  E₂ = + \(\frac { \sigma }{ 2{ \varepsilon }_{ 0 } } \widehat { r } \)    [due to negative plate]
\(\therefore\) New EFI at point P,                                   
E = E₁ + E₂ 
= \(\frac { \sigma }{ 2{ \varepsilon }_{ 0 } } \widehat { r } +\frac { \sigma }{ 2{ \varepsilon }_{ 0 } } \widehat { r } \)
\(\Rightarrow\)  E = \(\frac { \sigma }{ { \varepsilon }_{ 0 } } \widehat { r } \)
Thus, uniform electric field is produced between the two infinite parallel plane sheet of charge which is directed from positive plate to negative plate.

Let the equivalent capacitance of the series combination be C. Then,
\(\frac { 1 }{ C } =\frac { 1 }{ { { C }_{ 1 } } } +\frac { 1 }{ { C }_{ 2 } } +\frac { 1 }{ { C }_{ 3 } } =\frac { 1 }{ 10 } +\frac { 1 }{ 15 } +\frac { 1 }{ 30 } F\)  
\( =\frac { 3+2+1 }{ 30 } =\frac { 6 }{ 30 } =\frac { 1 }{ 5 } \Rightarrow C=5\mu \)
Charge stored on the combination is 
 q = CV = 5 x 60 = 300\(\mu C\) 
Since, the capacitor are in series, the charge on each capacitor is 300\(\mu C\).
The potential difference across the three capacitors are
 \({ V }_{ 1 }=\frac { q }{ { C }_{ 1 } } =\frac { 300 }{ 10 } =30V\)
\(\\ { V }_{ 2 }=\frac { q }{ { C }_{ 2 } } =\frac { 300 }{ 15 } =20V\)
\(\\ { V }_{ 3 }=\frac { q }{ { C }_{ 3 } } =\frac { 300 }{ 30 } =10V\)
Energy of the capacitors are
 \({ U }_{ 1 }=\frac { 1 }{ 2 } q{ V }_{ 1 }=\frac { 1 }{ 2 } \times 300\times 30=4500\mu J\)
\(\\ { U }_{ 2 }=\frac { 1 }{ 2 } q{ V }_{ 2 }=\frac { 1 }{ 2 } \times 300\times 20=3000\mu J\)
\(\\ { U }_{ 3 }=\frac { 1 }{ 2 } q{ V }_{ 3 }=\frac { 1 }{ 2 } \times 300\times 10=1500\mu J\)

Let E = 0 at a distance x to the right of \(-9\mu C\) charge.
Use k\({q_1\times1\over(8+x)^2}={kq_2\times 1\over x^2}\) 

Given q₁ = 2 \(\times\) 10^-7 C, q₂ = 3 \(\times\) 10^-7 C   r = 30 cm = 0.3 m
Therefore, Force of repulsion, F = 9 \(\times\) 10⁹ \(\times\) \(\frac { { q }_{ 1 }{ q }_{ 2 } }{ { r }^{ 2 } } \)
= \(9\times { 10 }^{ 9 }\times \frac { 2\times { 1 }0^{ -7 }\times 3\times { 10 }^{ -7 } }{ { (0.3) }^{ 2 } } =\frac { 54\times { 10 }^{ -5 } }{ 9\times { 10 }^{ -2 } } \)
 \(=\ 6\times { 10 }^{ -3 }N\)

The equipotential surfaces for an electric dipole are as shown below in the figure by dotted Iines.

Electric field is always perpendicular to an equipotential surface and as a result, work done in moving a charge between two points On an equipotential surface is zero.

The product of the magnitude of one of the point charges constituting an electric dipole and the separation between them is termed as electric dipole moment. It is a vector quantity.

Since, electric field is directed from higher electric potential to lower electric potential,
(i) Thus, V_B> V_A , so V_A - V_B will be negative
Further, d_AB = 2 cos 60° = 1m
\(\therefore\) V_A - V_B = - Ed_AB = (-10)(1) = -10 V
(ii) As, V_B > V_C so V_B - V_c will be positive
Further, d_BC = 2.0 m
\(\therefore\) V_B - V_c = (10) (2) = 20 V

Since, an electric dipole consists of two equal and opposite charges, the net charge on the dipole is zero.
Hence, the net electric flux coming out of the closed surface of the box or through the box is zero.

​​​​​​

In a pure semiconductors (Ge or Si) called intrinsic semiconductor the electrically conductivity is related by the electrons thermally excited from the valence band to the conduction band. There are equal number densities of free electrons and holes in conduction band and valence band of intrinsic semiconductor.
When a pure semiconductor of Ge or Si is doped with impurity atoms of valence three (like B or A1), some additional energy levels are created just above the upper energy level of valence band. Due to it, band gap of \(p\)-type semiconductor becomes smaller than intrinsic semiconductor. Also in \(p\)-type semiconductor, the number density of holes is more than that of electrons.
A \(p\)-type semiconductor is obtained when a trivalent atoms (Bor A1) are doped in pure semiconductor of Ge or Si. Here each doped trivalent atom shares its three valence electrons with the three atoms of Ge or Si and form bonds. While the fourth bond remains unbounded. Due to it, a hole is created. Since the impurity atoms and atoms of semiconductor are electrically neutral, hence \(p\)-type semiconductor is also neutral.

Work done in bringing q₂ from infinity to a point
\(=q_2 \times \frac{1}{4 \pi \epsilon_0} \frac{q_1}{r_{12}}\)
The potential energy of the system
\(=\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_{12}}\)
b) Total energy of the system
\(=q_1 V_1+q_2 V_2+\frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r_{12}}\)

The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an\(\alpha\) -particle and a gold nucleus is conserved. The system’s initial mechanical energy is E_t, before the particle and nucleus interact, and it is equal to its mechanical energy E_f when the (-particle momentarily stops. The initial energy E_t is just the kinetic energy K of the incoming \(\alpha\)- particle. The final energy E_f is just the electric potential energy U of the system. The potential energy U can be calculated from equation. 
Let d be the centre-to-centre distance between the (-particle and the gold nucleus when the (-particle is at its stopping point. Then we can write the conservation of energy E_t = E_f as
\(K=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{d}=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} d}\)
Thus the distance of closest approach d is given by
\(d=\frac{2 Z e^{2}}{4 \pi \varepsilon_{0} K}\)
The maximum kinetic energy found in\(\alpha\) -particles of natural origin is 7.7 MeV or 1.2 × 10^–12 J. Since 1/4 \(\pi\) E₀= 9.0 × 10⁹ N m²/C². Therefore with e = 1.6 × 10^–19 C, we have
\(d=\frac{(2)\left(9.0 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right)\left(1.6 \times 10^{-19} \mathrm{C}\right)^{2} \mathrm{Z}}{1.2 \times 10^{-12} \mathrm{~J}}\)
= 3.84 × 10^–16 Z m
The atomic number of foil material gold is Z = 79, so that d (Au) = 3.0 × 10^–14 m = 30 fm. (1 fm (i.e. fermi) = 10^–15 m.)
The radius of gold nucleus is, therefore, less than 3.0 × 10^–14 m. This is not in very good agreement with the observed result as the actual radius of gold nucleus is 6 fm. The cause of discrepancy is that the distance of closest approach is considerably larger than the sum of the radii of the gold nucleus and the \(\alpha\)-particle. Thus, the \(\alpha\)-particle reverses its motion without ever actually touching the gold nucleus.

(a) We can see from the figure that on the left face E and ΔS are parallel. Therefore, the outward flux is
φ_L= E.ΔS = – 200 \(\hat{i}\). ΔS
= + 200 ΔS, since\(\hat{i}\) . ΔS = – ΔS
= + 200 x \(\pi \) (0.05)² = + 1.57 N m² C^–1
On the right face, E and ΔS are parallel and therefore φ_R = E.ΔS = + 1.57 N m² C^–1
(b) For any point on the side of the cylinder E is perpendicular to ΔS and hence E.ΔS = 0. Therefore, the flux out of the side of the cylinder is zero
(c) Net outward flux through the cylinder
φ = 1.57 + 1.57 + 0 = 3.14 N m² C^–1

(d) The net charge within the cylinder can be found by using Gauss’s law which gives
q = ε₀φ
= 3.14 x 8.854 × 10^–12 C
= 2.78 x 10^–11 C

Reverse biasing

The VI characteristics are obtained by connecting the battery, to the diode, through a potentiometer (or rheostat). The applied voltage to the diode is changed. The values of current, for different values of voltage, are noted and a graph between V and I is plotted. The V-I characteristics, of a diode, have the form shown here.

(b) The circuit diagram, for the photodiode, is shown here.

The photodiode is illuminated by optical signal, whose photon energy is greater than the energy gap of the semiconductor used.
The electric field, at the junction, separates the electrons and holes and thus gives rise to an emf.
When an external load is connected, a (photo) current flows through it. The magnitude of this current is proportional to the intensity of light incident on the photodiode.

(a) Due to the diffusion of electrons and the holes, from their majority zone to minority zone, a layer of positive and negative space charge region on either side on the junction is formed. This is called the depletion region.
The loss of electrons, from n-region and gain of electrons by the p-region, causes a difference of potential across the junction. This tends to prevent the movement of charge carriers across the junction and is, therefore, termed as barrier potential.

For positive half cycle of input ac, one of the two diodes gets forward biased and conducts and output current is obtained across the load RL, For negative half cycle of input ac, the other diode
gets forward biased and thus output current is obtained due to it. Therefore, output is obtained for both the cycles of input ac.

(a) Electric dipole moment : The strength of an electric dipole is measured by the quantity electric dipole moment. Its magnitude is equal to the product of the magnitude of either charge and the distance between the two charges.
Electric dipole moment, p = q x d
It is a vector quantity.
In vector form, it is written as \(\overrightarrow{p}\) = q x 2 \(\overrightarrow{a}\)\(\hat{p}\)where direction of 2 \(\overrightarrow{a}\) \(\hat{p}\) is from negative charge to positive charge.
Electric field of dipole at points on the equatorial plane :

The magnitudes of the electric field due to the two charges +q and -q are given by,
\(E_{+q} = \frac{q}{4\pi\epsilon_0}\frac{1}{(r^2+a^2)}\)    ..........(i)
\(E_{-q} = \frac{q}{4\pi\epsilon_0}\frac{1}{(r^2+a^2)}\)  .............(ii)
\(\therefore\)  E_+q = E_-q
The direction of E_{+ q} and E_-q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
\(\therefore\) Total electric field E = - (E_+q+ E_{_q} ) cos q \(\hat{p}\)
[ Negative sign shows that field is opposite to \(\hat{p}\) ]
\(E = \frac{-2qa}{4\pi\epsilon_0(r^2+a^2)^{3/2}}\hat{p}\) ............(iii)
At large distances (r > > a), this reduces to
  \(E = \frac{2qa}{4\pi\epsilon_0r^3}\hat{p}\)  .................(iv)
  = 2qa\(\hat{p}\)
\(\therefore\)     =  \(\overrightarrow{p}\)
\(\therefore\) \(E = \frac{-\overrightarrow{p}}{4\pi\epsilon_0r^3}\) ( r > > a )
(b) Equipotential surface due to electric dipole:

The potential due to the dipole is zero at the line bisecting the dipole length

Here, \(R={2.4\over 2}=1.2m\)
\(\sigma=180.0 \mu C/m^2=180\times 10^{-6}Cm^{-2}\)
(i) Charge on the sphere q= \(4\pi R^2\sigma\)
\(=4\times 3.14\times(1.2)^2\times 180\times 10^{-6}\)
\(3.53\times 10^{-3}C\)
Electric flux, \(\phi_E={q\over \epsilon_o}={3.53\times10^{-3}\over8.85\times10^{-12}}\)

(d) : With the increase of temperature, the average energy exchanged in a collision increases and so more valence electrons can cross the energy gap, thereby increasing the electron-hole pairs. As in a semiconductor, conduction occurs mainly through electron-hole pairs, so conductivity increases with increase of temperature. Which in turn implies that the resistivity of a semiconductor decreases with rise in temperature.

(a): Capacitance of parallel plate capacitor is \(C=\frac{\varepsilon_{0} A}{d}\) Thus distance decreases and capacitance of capacitor increases.