\(\triangle\)Ie  = 7.89  x 10^-3 A
\(\triangle\)Ic = 7.8 x 10^-3 A
Now \(\alpha\)_a.c = \(\triangle\)Ic / \(\triangle\)Ie = 0.9886
We have, 
\(\beta\)_ac = \(\alpha\)_ac / (1- \(\alpha\)ac) = 86.72
Also, 
\(\beta\)ac = \(\triangle\)I_c/ \(\triangle\)I_b
\(\therefore\) \(\triangle\)I_b = Ic/\(\beta\)ac
\(\triangle\)I_b = (7.8 x 10^-3)/(86.72)
\(\triangle\)I_b = 89.94 x 10^-6A

e.m.f of the source , E = 1.5 
Voltage drop across the diode , V_d = 0.5 V
Maximum power rating of the diode 
I = (P / Vd)
I = 0.2A
Potential drop across resistance R
V = E - V_d
= 1 V
R = V/1 = 1/0.2 = 5 \(\Omega \)

(a) \(\frac { dN }{ dt } =-\lambda N\)
\(\int _{ { N }_{ 0 } }^{ N }{ \frac { dN }{ N } =\int _{ 0 }^{ t }{ -\lambda dt } } \) 
\(\left[ { log }_{ e }^{ N } \right] _{ { N }_{ 0 } }^{ N }=-\lambda \left[ t \right] _{ 0 }^{ t }\)
\(loge\frac { N }{ { N }_{ 0 } } =-\lambda t\)
\(N={ N }_{ 0 }{ e }^{ -\lambda t }\)
(b) (i) \(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Ne+{ e }^{ x }+\upsilon \)
Also accept,if a student does not identify the product nucleus and writes as
\(_{ 11 }^{ 22 }{ Na }\rightarrow _{ 10 }^{ 22 }Xe+{ e }^{ x }+\upsilon \)
Basic process  
\(p\rightarrow n+{ e }^{ + }+\upsilon \)
(ii) Isobar

Give ,carrier wave frequency
\({ f }_{ c }=20MHz=20\times { 10 }^{ 6 }Hz\)
Bandwidth required for modulation is
\({ 2f }_{ m }=3kHz=3\times { 10 }^{ 3 }Hz\)
\(\Rightarrow \ \ { f }_{ m }=\frac { 3\times { 10 }^{ 3 } }{ 2 } =1.5\times { 10 }^{ 3 }Hz\) s
Demodulation by a diode is possible,if the condition 
Thus,\(\frac { 1 }{ { f }_{ c } } =\frac { 1 }{ 20\times { 10 }^{ 6 }Hz } =0.5\times { 10 }^{ -7 }s\) .....(i)
and \(\frac { 1 }{ { f }_{ m } } =\frac { 1 }{ 1.5\times { 10 }^{ 3 }Hz } =0.7\times { 10 }^{ -3 }s\)    ......(ii)
Now,gain through all the option of R and C one by one,we get
(i)  \(RC=1k\Omega \times 0.01\mu F\)
\(\\ ={ 10 }^{ 3 }\Omega \times ({ 0.01 }\times { 10 }^{ -6 }F)={ 10 }^{ -5 }s\)
Here, condition \(\frac { 1 }{ { f }_{ c } } <\) is satisfied  
Hence, it can be demodulated.
(ii) \((RC=1k\Omega \times 0.01\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -8 }F={ 10 }^{ -4 }s)\)
Here, condition\(\frac { 1 }{ { f }_{ c } }\) < is satisfied
 Hence, it  can be demodulated
(iii) \(RC=10k\Omega \times 1\mu F={ 10 }^{ 4 }\Omega \times { 10 }^{ -6 }F={ 10 }^{ -2 }s\)
Here, condition\(\frac { 1 }{ { f }_{ c } } >RC,\) so this cannot be demodulated.

(i) By Malus law, intensity of emergent light from \({ P }_{ 2 }\) is \(I={ I }_{ \circ }cos^{ 2 }\theta \) , where \(\theta \) is the angle between \({ P }_{ 1 }and{ P }_{ 2 }\)|
When  \(\theta ={ 90 }^{ \circ }\) 
\(\\ I={ I }_{ \circ }\times 0\)
Intensity of emergent light, I = 0
(ii) Intensity of light from \({ P }_{ 3 }\)
 \(=\left( \frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta \right) \left[ cos^{ 2 }\left( { 90 }^{ \circ }-\beta \right) \right] \)

\(=\frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta sin^{ 2 }\beta =\frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta \)
\(As,\ \frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta =\frac { { I }_{ \circ } }{ 8 } (given)\)
\(so, \ \left( sin2\beta \right) ^{ 2 }=1 \ \Rightarrow 2\beta ={ 90 }^{ \circ } \ \Rightarrow \beta ={ 45 }^{ \circ }\)

de-Broglie matter wave equation,
\(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mK } } \ \left[ \because K=\frac { { P }^{ 2 } }{ 2m } \right] \)
where K is kinetic energy and m is a mass of the particle.
\(K=\frac { { h }^{ 2 } }{ 2m{ \lambda }^{ 2 } } \)   [ for same wavelength \( \lambda] \)
\(K\propto \frac { 1 }{ m }\)
\(\Rightarrow \ { K }_{ e }:{ K }_{ \alpha }:{ K }_{ p }=\frac { 1 }{ { m }_{ e } } :\frac { 1 }{ { m }_{ \alpha } } :\frac { 1 }{ { m }_{ p } } \)
where \({ m }_{ e },{ m }_{ p } \ and \ { m }_{ \alpha }\) are masses of electron, proton and \(\alpha \) -particle, respectively.
Also, \({ K }_{ e },k_{ p } \ and \ K_{ \alpha }\) are their respective kinetic energies.
\(\because \ m_{ \alpha }>m_{ p }>m_{ e }\)
\(\\ \Rightarrow m_{ \alpha }m_{ p }>m_{ e }m_{ \alpha }>m_{ e }m_{ p }\)
\(\\ { K }_{ e }>k_{ p }>K_{ \alpha }\)
(i) \(\alpha \) particle possess minimum kinetic energy
(ii) The electron has maximum kinetic energy. The magnifying power of an electron microscope is inversely related to the wavelength of radiation used. The Smaller wavelength of the electron beam in comparison to visible light increases the magnifying power of the microscope.

(i) The total charge of the capacitor remains conserved on introduction of dielectric slab. Also, the capacitance of capacitor increases to K times of original values. 
\(\therefore \\ \)  CV = C' V '= (KC) V' \(\Rightarrow V'=\frac { V }{ K } \) 
\(\therefore \\ \) New electric field,
\(E'=\frac { V' }{ d } =\left( { \frac { { V }/{ K } }{ d } } \right) =\left( \frac { V }{ d } \right) \frac { 1 }{ k } =\frac { E }{ K } \) 
​​\(\therefore \\ \)  On introduction of dielectric medium, new electric field E' becomes \(\frac { 1 }{ K } \) times of its original value (decrease).
(ii) The thickness of dieletric slab is \(\frac{d}{2}, \text { i.e. } t=\frac{d}{2}\)
The capacitance of a capacitor due to dielectric slab,
\(\begin{aligned} C & =\frac{\varepsilon_0 A}{d-t+\frac{t}{K}} \end{aligned}\)
\(\begin{aligned} =\frac{\varepsilon_0 A}{d-\frac{d}{2}+\frac{d}{2 K}}=\frac{2 \varepsilon_0 A}{d\left(1+\frac{1}{K}\right)} \end{aligned}\)

Let the distance between the plates be increased by a very small distance\(\triangle x\) = Force x Increased distance
= F.\(\Delta x\)     ...(i)  

Increase in volume of capacitor
= Area of plates x Increased distance
= A.\(\Delta x\)
u = Energy density = \(\frac { Energy }{ Volume } \)
Increase in energy = u x voulme = u.A.\(\Delta x\)   .....(ii)
As, Energy = Work done
F.\(\Delta x\) = u.A.\(\Delta x\) \([\because From \ Eqs.(i) \ and \ (ii)]\) 
 \(\Rightarrow\) F = u.A
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 }.A\quad [\because u=\frac { 1 }{ 2 } \varepsilon _{ 0 }{ E }^{ 2 } \ and \ E=\frac { V }{ d } ]\)
\(=\frac { 1 }{ 2 } \varepsilon _{ 0 }.\frac { { V }^{ 2 } }{ { d }^{ 2 } } .A=\left( \frac { { \varepsilon }_{ 0 }A }{ d } .A \right) \frac { V }{ d } \times \frac { 1 }{ 2 } \)
\(=\frac { 1 }{ 2 } .E.C.V=\frac { 1 }{ 2 } QE \ [\because C=\frac { { \varepsilon }_{ 0 }A }{ d } ,CV=Q]\)

Given, velocity of electromagnetic wave in vacum, c = 3 X 10⁸ m / s
Relative electric permeability,\(\varepsilon _{ r }\) = 2
and magnetic permeability, \(\mu _{ r }\) = 1
Since, velocity of electromagnetic wave in a medium can be calculated by
v = \(\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\varepsilon _{ r }\mu _{ 0 }\mu _{ r } } } =\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } \quad x \quad \sqrt { \mu _{ r }\varepsilon _{ r } } } \)
Where,
\(\frac { 1 }{ \sqrt { \varepsilon _{ 0 }\mu _{ 0 } } } =c \ \Rightarrow \ v \ = \ \frac { c }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)     ...(i)
Therefore, v \(=\frac { 3x10^{ 8 } }{ \sqrt { 2x1 } } \Rightarrow \) \(v=\frac { 3 }{ \sqrt { 2 } } \times10^{ 8 }\) m/s.

Microwaves are the electromagnetic waves of wavelength of the order of a few millimeters, which is less than those of T.V. signals. On account of smaller wavelength, the microwaves can be transmitted as beam signals in a particular direction, much better than radiowaves because microwaves do not spread or bend around the corners of any obstacle coming in their way. Therefore, microwaves are better carriers of signals than radio waves.

If electron is to move around the ion in circular path of radius r, then
Kinetic energy of electron,
\(\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { { e }^{ 2 } }{ 2\times 4\pi { \epsilon }_{ 0 }r } =\frac { { e }^{ 2 } }{ 8\pi { \epsilon }_{ 0 }r }\)
\(As \ \frac { 1 }{ 2 } { { mv }^{ 2 } }_{ max }=\frac { hc }{ \lambda } -\phi \ \therefore \ \frac { { e }^{ 2 } }{ 8\pi { \epsilon }_{ 0 }r } =\frac { hc }{ \lambda } -\phi \ or \ \frac { hc }{ \lambda } =\frac { { e }^{ 2 } }{ 8\pi { \epsilon }_{ 0 }r } +\phi =\frac { { e }^{ 2 }+8\pi { \epsilon }_{ 0 }r\phi }{ 8\pi { \epsilon }_{ 0 }r } \)
\( \lambda =\frac { 8\pi { \epsilon }_{ 0 }rhc }{ { e }^{ 2 }+8\pi { \epsilon }_{ 0 }r\phi } \)

In an ordinary atom, as a first approximation, we ignore the motion of the nucleus, being too heavy. In a positronium atom, a positron replaces proton of hydrogen atom. As electron and positron masses are equal, the motion of the positron cannot be ignored. We consider motion of electron and positron about their centre of mass. A detailed analysis (beyond the scope of this book) shows that formulae of Bohr model apply to positronium atom provided that we replace \({ m }_{ e }\) by what is known as reduced mass of the electron. For positronium, the reduced mass is \({ m }_{ e }/2.\) In the transition n = 2 to n = 1, the wavelength of radiation emitted is double than that of the corresponding radiation emitted for a similar transition in hydrogen atom, which has a wavelength of \(1217\mathring { A } \) ; and hence is equal to \(2\times 1217=2434\mathring { A } .\) This radiation lies in the ultra-violet part of the electromagnetic spectrum.

(i) Electric field at x = 0
\(E={1\over 4\pi \epsilon_o}[{q\over 1^2}+{q\over 2^2}+{q\over 4^2}+....]\)
\(={q\over 4\pi\epsilon_o}[{1\over 1-1/4}]={q\over 3\pi\epsilon_o}\)
(ii) If consecutive charges have opposite signs, then at x = 0, 
Electric field, \(E'={q\over 4\pi\epsilon_o}[{1\over 1^2}-{1\over 4}+{1\over 16}-{1\over 64}+....]^\times\)
\(E'={q\over 4\pi\epsilon_o}[{1\over 1-(-1/4)}]={q\over 5\pi\epsilon_o}\) 

\( { I }_{ v }=\frac { { E }_{ v } }{ Z } =\frac { 130 }{ 2 } \ \ ...(i)\)
\(Since \ \ Z=\sqrt { { R }^{ 2 }+{ X }_{ L }^{ 2 } } =\sqrt { { \left( 12 \right) }^{ 2 }{ +\left( 2\pi vL \right) }^{ 2 } }\)
\(=\sqrt { { \left( 12 \right) }^{ 2 }+{ \left( 2\pi \times 50\times \frac { 0.05 }{ \pi } \right) }^{ 2 } } \)
\(or \ Z=\sqrt { 144+{ \left( 5 \right) }^{ 2 } } =13\Omega \)
\(So \ \ { I }_{ v } \ =\frac { 130 }{ 13 } =10 \ A\)
\(\\ And \ \ { E }_{ L }={ X }_{ L }I=5\times 10=50 \ V\)