(i) \((A+B)(A+B)=AA+AB+BA+BB\)     \((\because AA=A)\) 
\(=A+AB+B\) 
= A+ AB +B = A(1 + B) + B    \((\because AB+AB=AB)\)
= A + B   \((\because 1+B=1)\) 
(ii) \(A(\bar{A}+B)\)\(=A \bar{A}+A B=AB\)      \((\because A\bar{A}=0)\)
(iii) (A + B) (A + C) =  AA + AC + BA + BC
= A + AC + BA + BC
= A (1 + C) + BA + BC
= A + BA + BC
= A(1 + B) + BC   \((\because 1+C=1)\)
= A + BC  \((\because 1+B=1)\)

In photoelectric effect, a single photon interacts with a single electron. As individual photo has energy (2.5 eV) which is less than work function, hence emission of electron will not take place.

i) The threshold wavelength is given by
\({ \lambda }=\frac { hc }{ { \phi }_{ 0 } } =\frac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 4.65\times 1.6\times 10^{ -19 } } \)
= 2672 \(\mathring { A }\)
ii) Energy of the photon of wavelength 2200 \(\mathring { A }\) is
E = \(\frac { hc }{ \lambda } =\frac { 6.626\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ 2200\times 10^{ -10 } } \)
= 9.035 x 10^-19 J = 5.65 eV
We know that kinetic energy of fastest photo electron is
K_max = hv - ϕ₀ = 5.65 - 4.65
= 1 eV
From equation (7.3), K_max = eV₀
V₀ = \(\frac { { K }_{ max } }{ e } =\frac { 1\times 1.6\times { 10 }^{ -19 } }{ 1.6\times { 10 }^{ -19 } } \)
Therefore, stopping potential = 1 V

Energy of the incident photon is
E = hv = \(\frac { hc }{ \lambda } \) (in joules)
E = \(\frac { hc }{ \lambda e } \) (in eV)
Substituting the known values, we get
E = \(\frac { 6.626\times { 10 }^{ -34 }\times 3\times 10^{ 8 } }{ 300\times { 10 }^{ -9 }\times 1.6\times { 10 }^{ -19 } } \)
E = 4.14 eV
The work function of silver = 4.7 eV. Since the energy of the incident photon is less than the work function of silver, photoelectrons are not observed in this case.

Brewster’s law, tan i_p = n
For glass, tani_p = 1.5 ; i_p = tan^-11.5 ; i_p= 56.3^o
For water, tani_p= 1.33; i_p= tan^-1 = tan^-1 1.33; i_p = 53.1^o

Step 1: AC (1 + B) = AC.1 [OR law-2]
Step 2: AC . 1 = AC [AND law – 2]
Therefore, AC + ABC = AC
Thus the Boolean identity is proved.
Circuit Description

α = \(\frac{I_C}{I_E}\)
I_C = α I_E = 0.95 x 1 = 0.95 mA
I_E = I_B + I_C
∴ I_B = I_E - I_C = 1 - 0.95 = 0.05 mA

The refractive index of vacuum, n₁ = 1
The wavelength in vacuum, λ₁ = 5893 Å.
The speed in vacuum, c = v₁ = 3 x 10⁸ m s^–1
The refractive index of water, n₂ = 1.33
The wavelength of light in water, λ₂
The speed of light in water, v₂
(a) The equation relating the wavelength and refractive index is,
\(\cfrac { { \lambda }_{ 1 } }{ \lambda _{ 2 } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) 
Rewriting, \({ \lambda }_{ 2 }=\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \times { \lambda }_{ 1 }\)
Substituting the values,
 \({ \lambda }_{ 2 }=\cfrac { 1 }{ 1.33 } \times 5893\overset { o }{ A } =4431\overset { o }{ A } \) 
\({ \lambda }_{ 2 }=4431\overset { o }{ A } \) 
(b) The equation relating the speed and refractive index is,
\(\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Rewriting, \({ v }_{ 2 }=\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \times { v }_{ 1 }\) 
Substituting the values,
\({ v }_{ 2 }=\cfrac { 1 }{ 1.33 } \times 3\times { 10 }^{ 8 }=2.256\times { 10 }^{ 8 }\) 
v₂ = 2.256 x 10⁸ ms^-1
(c) Frequency of light in vacuum is,
\({ v }_{ 1 }=\cfrac { c }{ { \lambda }_{ 1 } } \) 
Substituting the values,
\({ v }_{ 1 }=\cfrac { 3\times { 10 }^{ 8 } }{ 5893\times { 10 }^{ -10 } } =5.091\times { 10 }^{ 14 }Hz\)
Frequency of light in water is, \({ v }_{ 2 }=\cfrac { v }{ { \lambda }_{ 2 } } \)
Substituting the values,
 \(\\ { v }_{ 2 }=\cfrac { 2.256\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 4431\times { 10 }^{ -10 } } =5.091\times { 10 }^{ 14 }Hz\)
The results show that the frequency remains same in all media.

235 g of \(_{ 92 }^{ 235 }{ U }\) has 6.02 x 10²³ atoms. In one gram of \(_{ 92 }^{ 235 }{ U }\), the number of atoms is equal to \(\frac { 6.02\times { 10 }^{ 23 } }{ 235 } =2.56\times { 10 }^{ 21 }\)
So the number of atoms in 1 kg of \(_{ 92 }^{ 235 }{ U }\) = 2.56 x 10²¹ x 1000 = 2.56 x 10²⁴
Each \(_{ 92 }^{ 235 }{ U }\) nucleus releases 200 MeV of energy during the fission. The total energy released by 1kg of \(_{ 92 }^{ 235 }{ U }\) is
Q = 2.56 x 10²⁴ x 200MeV = 5.12 x 10²⁶ MeV
In terms of joules,
Q = 5.12 x  10²⁶ x 1.6 x 10^-13 J = 8.192 x 10¹³ J
In terms of Kilowatt hour,
Q = \(\frac { 8.192\times { 10 }^{ 13 } }{ 3.6\times { 10 }^{ 6 } } =2.27\times { 10 }^{ 7 }\) kWh

To get the time interval in terms of half life, \(n=\frac { t }{ { T }_{ 1/2 } } =\frac { 22,920 \ yr }{ 5730 \ yr } =4\)
The number of nuclei remaining undecayed after 22,920 years,
\(N={ \left( \frac { 1 }{ 2 } \right) }^{ n }{ N }_{ 0 }={ \left( \frac { 1 }{ 2 } \right) }^{ 4 }\times 10,000\)
N = 625

From example, we found that the BE of  \(_{ 2 }^{ 4 }{ He }\) = 28.33 Mev
Binding energy per nucleon = \(\overline{B \cdot E}\) = \(28.33 \mathrm{MeV} / 4 \simeq 7 \mathrm{MeV}\).

R = R₀A^\(\frac13\)
R = 1.2 x 10^{−15 x} (197)^\(\frac13\) = 6.97 x 10⁻¹⁵ m
Or R = 6.97 F

\(T_{1 / 2}=1600 \text { years }=1600 \times 365 \times 24 \times 60 \times 60 s\)
R = 1 curie = 3.7 x 10¹⁰ Bq, Show that m = 1g
R = λN
Number of atoms Present, N = \(\frac{\mathrm{R}}{\lambda}=\frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2}\)
Mass of 6.023 x 10²³ atoms of  \({ }_{88}^{226} R a=226 g\)
Mass of 1 atom of  \({ }_{88}^{226} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \mathrm{~g}\)
Mass of N atoms of \({ }_{88}^{{ }{266}} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \times \mathrm{Ng}\)
Mass of N atoms of \({ }_{88}^{226} \mathrm{Ra}(\mathrm{m})=\frac{226}{6.023 \times 10^{23}} \times \frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2} \mathrm{~g}\)
\(\mathrm{m}=\frac{226}{6.023 \times 10^{23}} \times \frac{3.7 \times 10^{10}}{0.6931} \times 1600 \times 365 \times 24 \times 60 \times 60 \mathrm{~g}\)
m = 1.01 g

Energy produced per second in reactor = 1 W = 1 J/s
Energy produced per fission = 200 MeV = 200 x 10⁶ x 1.6 x 10^-19 J
= 3.2 x 10¹¹ J
Number of fissions per second required \(=\frac{\text { Energy produced per second in reactor }}{\text { Energy produced per fission }} \)
\(=\frac{1}{3.2 \times 10^{11}}=\frac{10 \times 10^{10}}{3.2}=3.125 \times 10^{10} \)
Number of fissions per second = 3.125 x 10¹⁰ 

The total distance d between the transmitting and receiving antennas will be the sum of the individual distances of coverage.
d = d₁ + d₂
\(=\sqrt { 2R{ h }_{ 1 } } +\sqrt { 2{ Rh }_{ 2 } } \)
\(=\sqrt { 2R } \left( \sqrt { { h }_{ 1 } } +\sqrt { { h }_{ 2 } } \right) \)
\(=\sqrt { 2\times 6.4\times { 10 }^{ 6 } } \times (\sqrt { 40 } +\sqrt { 30 } )\)
\(=16\times { 10 }^{ 2 }\sqrt { 5 } \times (6.32+5.48)\)
= 42217 m = 42.217 km

\(\begin{array}{l} Y=(\bar{A}+\bar{B}) \cdot C \\ Y=(\bar{A} \cdot \bar{B}) \cdot C \end{array}\)         (Using De-Morgan's 1" theorem)      
Y =  ABC

After the Emission of \(2\alpha\) particles,
\({ }_{92} \mathrm{U}^{235} \rightarrow{ }_{\mathrm{88} } \mathrm{Ra}^{227}+2_2 \mathrm{He}^4\) 
After the Emission of \(3\beta\)  particles,
\({ }_{88} \mathrm{Ra}^{227} \rightarrow{ }_{91} \mathrm{~Pa}^{227^*}+3_{-1} \mathrm{e}^0\)
After the Emission of \(2\gamma\) particles,
\({ }_{91} \mathrm{~Pa}^{227^*} \rightarrow{ }_{91} \mathrm{~Pa}^{227}+2 \gamma\)
Therefore, the resulting
Atomic Number = 91
Mass Number = 227 and the Radio active Element is Protactinium

An alpha particle contains 2 protons and 2 neutrons. Therefore, the mass M of the alpha particle is 4 times that of a proton (m_p) (or a neutron) and its charge q is twice that of a proton (+e).
The de Broglie wavelength associated with it is
\(\lambda=\frac { h }{ \sqrt { 2MqV } } =\frac { h }{ \sqrt { 2\times (4m_{ p })\times (2e)\times V } } \)
\(=\frac { 6.626\times { 10 }^{ -34 } }{ \sqrt { 2\times 4\times 1.67\times 10^{ -27 }\times 2\times 1.6\times { 10 }^{ -19 }\times 400 } } \)
\(=\frac { 6.626\times 10^{ -34 } }{ 4\times 20\times { 10 }^{ -23 }\sqrt { 1.67\times 1.6 } } \) = 0.00507 \(\mathring { A }\)

\(\lambda=6000Å=6000 \times 10^{-10} \mathrm{~m} ; \mathrm{V}=0.8 \mathrm{v} \)
\(\mathrm{k} \cdot \mathrm{E}=\mathrm{hv}-\phi \)
\(\mathrm{eV}_o=\mathrm{hv}-\phi=\frac{\mathrm{hc}}{\lambda}-\phi \)
\((i) v=\frac{c}{\lambda}=\frac{3 \times 10^{8}}{6000 \times 10^{-10}}=5 \times 10^{14} \mathrm{~Hz} \)
\((ii)\ \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10}}=3.313 \times 10^{-19} J\)
\(\mathrm{E}=\frac{3.313 \times 10^{-19}}{1.6 \times 10^{-19}}=2.07 \mathrm{eV} \)
\((iii) \ \mathrm{E}=\mathrm{hv}-\mathrm{W}\Rightarrow\mathrm{W}=\mathrm{h} v-\mathrm{E} \)
\(\mathrm{E}=\mathrm{eV_o}=1.6 \times 10^{-19} \times 0.8=1.2 8 \times10^{-19}J\)
\(\mathrm{hv}=6.626 \times 10^{-34} \times 5 \times 10^{14}=3.313 \times 10^{-19}J \)
\(\mathrm{~W}=\frac{(3.313-1.28) \times 10^{-19}}{1.6 \times 10^{-19}}=1.270 \mathrm{eV} \)
W = 1.27 eV
\((iv) \ \mathrm{W}=\mathrm{h} \mathrm{v}_{0} \)
\(v_{0}=\frac{W}{h}=\frac{2.033 \times 10^{-19}}{6.626 \times 10^{-34}}=3.07 \times 10^{14} \mathrm{~Hz} \)
\((v)\ \mathrm{E}=\mathrm{eV}_o=\frac{0.8 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=\mathbf{0 . 8} \mathrm{eV}\)

\(v= 9 \times 10^{14} \mathrm{~Hz} ; \mathrm{V}=8 \times 10^{5} \mathrm{~m} / \mathrm{s} \)
\(\mathrm{E}= \mathrm{h}-\mathrm{h} v_{o} \Rightarrow v_{o}=\frac{\mathrm{h}v-\mathrm{E}}{\mathrm{h}}=\mathrm{V}-\frac{\mathrm{E}}{\mathrm{h}}=\mathrm{V}-\frac{\frac{1}{2} \mathrm{mv}^{2} }{h}\)
\(v_s={9 \times 10^{14}-\frac{\left[\frac{1}{2} \times 9.1 \times 10^{-31} \times\left(8 \times 10^{5}\right)^{2}\right]}{6.626 \times 10^{-34}}}=4.605 \times 10^{14} \)
\( v_{o} \simeq 4.6 \times 10^{14} \mathrm{~Hz} \)

\(v=10^{14} \mathrm{~Hz} ; \mathrm{E}=19.86 \mathrm{~J} \)
\(E=nhv \Rightarrow n=\frac{E}{hv}\)
\(=\frac{19.86}{6.626 \times 10^{-34} \times 10^{14}}=2.99 \times 10^{20} \)
\(\mathrm{n} \simeq 3 \times 10^{20} \)

\(\text{ longitudinal magnifcation}(m_l)=\frac { length\ of\ image\left( l' \right) }{ length\ of\ object\left( l \right) } \)
Given: length of object, \(l=\cfrac { f }{ 3 } \) 
For the given condition, the image formation is shown in the figure.

Let, l' be the length of the image, then
\(m=\cfrac { l' }{ l } =\cfrac { l' }{ f/3 } \) (or) \(l=\cfrac { m_lf }{ 3 } \) 
Image of one end coincides with the object. Thus, the coinciding end must be at center of curvature.
 \(u_B=u_A-\cfrac { f }{ 3 } =2f-\cfrac { f }{ 3 } =\cfrac { 5f }{ 3 } \)
\(v_B=u_B+l+l'\) 
\(v_b =\cfrac { 5f }{ 3 } +\cfrac { f }{ 3 } +\cfrac { mf }{ 3 } =\cfrac { f(6+m) }{ 3 } \)
 Mirror equation,\(\cfrac { 1 }{ v } +\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
\(\cfrac { 1 }{ -\left( \cfrac { f(6+m_l) }{ 3 } \right) } +\cfrac { 1 }{ -\left( \cfrac { 5f }{ 3 } \right) } =\cfrac { 1 }{ -f } \)
After simplifying,
 \(\cfrac { 3 }{ f(6+m_l) } +\cfrac { 3 }{ 5f } =\cfrac { 1 }{ f } ;\cfrac { 3 }{ (6+m_l) } =\cfrac { 2 }{ 5 } \) 
\(6+m_l=\cfrac { 15 }{ 2 } ;m_l=\cfrac { 15 }{ 2 } -6\) 
\(m_l=\cfrac { 3 }{ 2 } =1.5\) 

\(P_{a}=\frac{1}{f_{a}}=\left(\frac{\mu_{g}}{\mu_{a}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ..(1)
\(P_{w}=\frac{1}{f_{w}}=\left(\frac{\mu_{g}}{\mu_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)  ...(2)
\(p_{a}=5 D, f_{w}=-100 \text { (Diverging lens) } \)
\(\mu_{\mathrm{g}}=1.5, \mu_{\mathrm{a}}=1 \)
\(5=(1.5-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)   ...(3)
\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)=\frac{5}{0.5}=\frac{50}{5}=10 \)
\(\frac{1}{f_{w}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)    ...(4)
\(\frac{-1}{100 \times 10^{-2}}=\left(\frac{1.5}{n_{w}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right) \)
\(-1=\left(\frac{1.5}{n_{w}}-1\right)(10) \Rightarrow \frac{-1}{10}=\frac{1.5}{n_{w}}-1 \)
\(\frac{1.5}{n_{w}}=\frac{-1}{10}+1 \)
\(\frac{1.5}{n_{w}}=\frac{9}{10} \)
\(n_{w}=\frac{1.5 \times 10}{9}=\frac{15}{9}=\frac{5}{3} \)
\(n_{w}=\frac{5}{3} \)

Decayed = 60 %
Left undecayed = 40 %(ie) \(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\frac{40}{100} \)
\(\mathrm{~T}_{\frac{1}{2}}=3.8 \text { days } \)
\(\mathbf{N}=\mathrm{N}_{0} \mathrm{e}^{-\lambda t} \)
\(\frac{\mathrm{N}}{\mathrm{N}_{0}}=\mathrm{e}^{-\lambda t} \)
\(\frac{40}{100}=\mathrm{e}^{-\lambda t} \Rightarrow \frac{100}{40}=2.5=\mathrm{e}^{\lambda t} \)
\(\therefore \mathrm{e}^{\lambda t} \) = 2.5
Taking log on both sides
\(\lambda t=\ln [2.5]=2.3026 \times \log (2.5)=2.3026 \times 0.3974 \)
\(t=\frac{0.9163}{\lambda}=\frac{0.9163}{0.6931} \times T_{1 / 2} \)
\(t=1.322 \times 3.8=5.022 \text { days } \)

\(E=hv,v=\frac { E }{ h } =\frac { 66.3\times 1.6\times { 10 }^{ -19 } }{ 6.63\times { 10 }^{ -34 } } \)
= 16 x 10¹⁵ hHz = 1.6 x 10¹⁶