Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
(a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
\(r_{f b}=\Delta V / \Delta I=0.1 \mathrm{~V} / 10 \mathrm{~mA}=10 \ \Omega\)
(b) From the curve at V = –10 V, I = –1 \(\mu\) A,
Therefore,
\(r_{r b}=10 \mathrm{~V} / 1 \mu \mathrm{A}=1.0 \times 10^{7} \ \Omega\) 

(i) Since, Q has greater negative intercept, it will have greater \(\phi\) (work function) and hence higher threshold frequency.
(ii) To know work function of Q, we put
V = 0 in the following equation.
\(\begin{array}{rlrl} V =\frac{h v}{e}-\frac{\phi}{e} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & 0 & =\frac{h v}{e}-\frac{\phi}{e} \Rightarrow \phi=h v \\ \end{array}\)
\(\begin{array}{rlrl} \therefore & \phi & =6.6 \times 10^{-34} \times 6 \times 10^{14} \mathrm{~J} \end{array}\)
\(=\frac{6.6 \times 6 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}=2.5 \mathrm{eV}\)
(iii) From the equation, \(v \lambda=c\)
\(\begin{aligned} \Rightarrow \quad \lambda & =\frac{c}{v}=\frac{3 \times 10^8}{8 \times 10^{14}}=\frac{30}{8} \times 10^{-7} \mathrm{~m} \end{aligned}\)
\(\begin{aligned} =\frac{30}{8} \times 10^3 \times 10^{-10} \mathrm{~m} \end{aligned}\)
\(\begin{aligned} =\frac{30}{8} \times 10^3 \end{aligned}\) \(\overset{\circ}{A}\)= 3750 \(\overset{\circ}{A}\)
Energy \(=\frac{12375}{\lambda(\overset{\circ}{A})} \mathrm{eV}=\frac{12375}{3750} \mathrm{eV}=33 \mathrm{eV}\)
\(\therefore\) Maximum KE of emitted electron = 33 - 2.5 eV
= 0.8 eV

(i) From Einstien's photoelectric equation, we have,
K_max = hf\({ \phi }_{ 0 }\)
where, K_max is maximum kinetic energy of the photoelectrons, G>o is work function and hf is energy of the incident photon.
As K_max ≥ 0
So, \(hf-{ \phi }_{ 0 }\ge 0 \ or \ f\ge \frac { { \phi }_{ 0 } }{ h } \)
Thus, photoemission occurs, when frequency is greater than threshold frequency, \({ f }_{ 0 }\ge \frac { { \phi }_{ 0 } }{ h } \)
Energy of the incident radiation of wavelength \(\lambda \),
\(E=\frac { hc }{ \lambda } =\frac { \left( { 6.63\times 10 }^{ -34 } \right) \times \left( { 3\times 10 }^{ 8 } \right) }{ 3300\times { 10 }^{ -10 }\times 1.6\times { 10 }^{ -19 } } =3.76eV\)
This energy of the incident radiation is greater than the work function of Na and K but less than those of Mo and Ni. so photoelectric emission will occur only in Na and K metals and not in Mo and Ni.
If the laser is brought closer, the intensity of incident radiation increases. This does not affect the result regarding Mo and Ni metals, while photoelectric current from Na and K will increase in proportion to intensity.

Given, Bohr's radius, \({ r }_{ 1 }=5.3\times 10^{ -11 }\)
We know that, \({ r }_{ n }=n^{ 2 }{ r }_{ 1 }\)
Let  be radius of the orbit for n = 2
Therefore, \(r_{ 2 }=(2)^{ 2 }\times 5.3\times 10^{ -11 }\quad =2.12\times 10^{ -10 }m\)
(ii) Given, total energy of an electron in second excited state,
E = 1.51 eV
(a) Kinetic energy of electron is equal to negative of the total energy
\(\Rightarrow \)  K = -E = -(-1.51) = 1.51 eV
(b) Potential energy of electron is equal to negative of twice of its kinetic energy
\(\Rightarrow \) U = -2K = -2 x 1.51 = -3.02 eV

Einstein's photoelectric equations and its features
According to the photoelectric equation,
\({ K }_{ max }=\frac { 1 }{ 2 } { mv }_{ max }^{ 2 }=hv-{ \phi }_{ 0 }\)
\({ K }_{ max }=\frac { hc }{ { \lambda }_{ 1 } } -{ \phi }_{ 0 }\ ........... (i)\)
Let the maximum kinetic energy for the incident radition (of wavelength \({ \lambda }_{ 2 }\)) be \({ K }_{ max }^{ ' }\).
\(\Rightarrow \ { K }_{ max }^{ ' }=\frac { hc }{ { \lambda }_{ 2 } } -{ \phi }_{ 0 } ........... (ii)\)
From Eqs. (i) and (ii), we get
\(\frac { hc }{ { \lambda }_{ 0 } } -{ \phi }_{ 0 }=2\left( \frac { hc }{ { \lambda }_{ 1 } } -{ \phi }_{ 0 } \right) \quad \quad [\because { K }_{ max }^{ ' }=2\quad { K }_{ max }]\)
\( \Rightarrow \quad { \phi }_{ 0 }=hc(\frac { 2 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } )\)
\( \Rightarrow h{ v }_{ 0 }=hc(\frac { 2 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } )\)
\( \frac { c }{ { \lambda }_{ 0 } } =c(\frac { 2 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } )\)
\(\\ \Rightarrow \frac { 1 }{ { \lambda }_{ 0 } } =(\frac { 2 }{ { \lambda }_{ 1 } } -\frac { 1 }{ { \lambda }_{ 2 } } )\)
\(\Rightarrow { \lambda }_{ 0 }=(\frac { { \lambda }_{ 1 }{ \lambda }_{ 2 } }{ 2{ \lambda }_{ 2 }-{ \lambda }_{ 1 } } )\)

\(Here \ { \phi }_{ 0 }=1.8eV,\lambda =4500 \ A=4.5\times { 10 }^{ -7 }m\)
(i) Max K.E of emitted photoelectron is
\({ K }_{ max }=\frac { hc }{ \lambda } -{ \phi }_{ 0 }\)
\( =\frac { (6.6\times { 10 }^{ -34 })(3\times { 10 }^{ 8 }) }{ 4.5\times { 10 }^{ -7 } } -1.8\times 1.6\times { 10 }^{ -19 }\)
\( =4.4\times { 10 }^{ -19 }-2.88\times { 10 }^{ -19 }=1.52\times { 10 }^{ -19 }J\)
(ii)Max. velocity of emitted photoelectron
\({ v }_{ max }=\sqrt { \frac { { 2K }_{ max } }{ m } } =\sqrt { \frac { 2\times 1.52\times { 10 }^{ -19 } }{ 9.1\times { 10 }^{ -31 } } }\)
\( =5.78\times { 10 }^{ 5 }{ ms }^{ -1 }\)
(iii) The kinetic energy of the emitted photoelectron is an incident of the intensity of the incident light. Hence, if the intensity of incident light is doubled the max. K.E of the emitted photoelectron electrons remains unchanged.

\((i) \ -1.7eV; \ 0.85 \ eV; \ (ii) \ 3\times { 10 }^{ 15 }Hz\)
Here,   \({ E }_{ 1 }=-13.6\quad eV\)
For third excited state, n = 4
\(\therefore \ { E }_{ 4 }=\frac { -13.6 }{ { 4 }^{ 2 } } =-0.85 \ eV\)
\( \therefore K.E=-{ E }_{ 4 }=0.85eV\)
\( P.E=-2(K.E)=-2(0.85)eV=-1.70eV\)
Energy emitted, \(\Delta E={ E }_{ 4 }-{ E }_{ 1 }\)
\(hv=-0.85-(-13.6)eV=12.75 \ eV\)
\(v=\frac { 12.75\times 1.6\times { 10 }^{ -19 } }{ 6.6\times { 10 }^{ -34 } } =3\times { 10 }^{ 15 }Hz\)

Neutron separation energy \({ S }_{ n }\) of a nucleus \(\\ \begin{matrix} A \\ X \end{matrix}X\) 
Given by \({ S }_{ n }=[{ m }_{ N }(\begin{matrix} A-1 \\ Z \end{matrix}X)+{ m }_{ n }-{ m }_{ N }(\begin{matrix} A \\ X \end{matrix}X)]{ c }^{ 2 }\).
Adding and subtracting the term \({ Zm }_{ C }\) in the bracket above ignoring mass defects due to electronic binding energies we get \({ S }_{ n }\) in terms of atomic masses.

\({ S }_{ n }=[{ m }_{ N }=(\begin{matrix} A-1 \\ Z \end{matrix}X)+{ m }_{ n }-{ m }(\begin{matrix} A \\ Z \end{matrix}X)]{ c }^{ 2 }\)
\({ S }_{ n }(\begin{matrix} 41 \\ 20 \end{matrix}X)=[39.96259+1.008665-40.962278)]u,\)
\(=[40.971255-40.962278]931.5\)
\(=0.008978 \times 931.5\)
\(=8.363 \ MeV.\)
\(\\ Neutron \ separation \ energy \ for \ \begin{matrix} 27 \\ 13 \end{matrix}Al\)
\({ S }_{ n }(\begin{matrix} 27 \\ 13 \end{matrix}Al)=[{ m }_{ N }(\begin{matrix} 26 \\ 13 \end{matrix}Al)+{ m }_{ n }-{ m }_{ N }(\begin{matrix} 27 \\ 13 \end{matrix}Al)]{ c }^{ 2 }\)
\({ S }_{ n }(\begin{matrix} 27 \\ 13 \end{matrix}Al)=[25.986895+1.0008665-26.981541]u\)
\( =[26.99556-26.981541]931.5\)
\( =0.014019 \times 931.5\)
\(=13.06 \ MeV.\)

Circuit diagram of a p-n junction diode in
(i) Forward biasing

(ii) Reverse biasing

I-V characteristics of p-n junction diode

Given, \(\lambda=412.5 \mathrm{nm}=412.5 \times 10^{-9} \mathrm{~m}\)
\( \therefore \ E =\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{412.5 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV}\)
\( =3.01 \mathrm{eV}\)
From the given question, work function (\(\phi\))of the following metals are given as
\(\mathrm{Na} \rightarrow 1.92, \mathrm{~K} \rightarrow 2.15 \)
\(\mathrm{Ca} \rightarrow 3.20, \mathrm{Mo} \rightarrow 4.17 \)
As the given energy is greater than the work function of Na and K only, hence these metals shows photoelectric emission.

(i) For hydrogen atom,
E₁ = -13.6 eV
E₂ = -3.4 eV
E₃ = -1.51 eV
E₄ = -0.85 eV
h = 6.63x10^-34 Js;
c = 3 x 10⁸ ms^-1
Photon Energy = \(\frac{hc}{\lambda}\)
=\(\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{496\times 10{-9}\times 1.6\times10^{-19}}\)
= 2.5 eV
This equals (nearly) the difference (E₄- E₂).
Hence the required transition is (n = 4) to ( n = 2)
(ii) The transition n = 4 to n = 3 corresponds to emission of radiation of maximum wavelength.
It is so because this transmission gives out the photon of least energy.

The photoelectric current is directly proportional to the intensity of incident radiation. Energy of photoelectrons or cut-off potential depends on frequency of incident radiation. Curves, a and b have got same cut-off potential, so for these two curves frequencies will be same.

According to the question,
 
D₂ is in reverse bias, so it acts as open circuit
\(\begin{aligned} R_{\mathrm{eq}} & =2+1=3 \Omega \end{aligned}\)
\(\begin{aligned} I & =\frac{V}{R_{\mathrm{eq}}}=\frac{6}{3}=2 \mathrm{~A} \end{aligned}\)

Given, for the first condition, \(\lambda\) = 600 nm
For the second condition, \(\lambda\)' = 400nm
K'_max = 2K_max
Here, \(K_{\max }^{\prime}=\frac{h c}{\lambda}-\phi \Rightarrow 2 K_{\max }=\frac{h c}{\lambda^{\prime}}-\phi_0\)
\( \Rightarrow 2\left(\frac{1240}{600}-\phi\right) \approx\left(\frac{1240}{400}-\phi\right)[\because h c \approx 1240 \mathrm{eV}-\mathrm{nm}] \\ \Rightarrow \phi=\frac{1240}{1200}=1.03 \mathrm{eV} \)

As we know, de-Broglie wavelength \(\lambda=\frac{h}{p}=\frac{h}{\sqrt{2 m K}}\)
\(K_1=\frac{h^2}{2 m \lambda_1^2}\)
If according of the questions
\( K_2 =\frac{K_1}{4} \\ K_2 =\frac{h^2}{2 \lambda_2^2} \\ \frac{K_1}{4} =\frac{h^2}{2 m \lambda_2^2} \)
From Eqs. (i) and (ii), we get
\( \frac{\frac{K_1}{4}}{K_1}=\frac{h^2}{2 m \lambda_2^2} \times \frac{2 m \lambda_1^2}{h^2} \\ \frac{1}{4}=\frac{\lambda_1^2}{\lambda_2^2} \\ \Rightarrow \quad \frac{\lambda_1}{\lambda_2}=\frac{1}{2} \)
Hence, wavelength of the particle double the wavelength when kineettiic energy is 1/4^th.

The energy of blue light \((h v)_{b l u e}\) is greater than the energy of red light. \((h v)_{r e d}\) In photoelectric emission, max K.E. of emitted electron
\(=h v-\phi_0, \text { i.e., } \max K . E \alpha(h v)\)
So, K.E of emitted electrons is more with blue light than that of red light. 

At the distance of closest approach \(\left( { r }_{ 0 } \right) \),
\(K{ E }_{ \alpha }=\frac { \left( Ze \right) \left( 2e \right) }{ 4\pi { \epsilon }_{ 0 }{ r }_{ 0 } } \)  and \(K{ E }_{ p }=\frac { \left( Ze \right) \left( e \right) }{ 4\pi { \epsilon }_{ 0 }{ r }_{ 0 } }\).
Clearly, \(K{ E }_{ p }=\frac { 1 }{ 2 } K{ E }_{ \alpha } \)

(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.

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.

(i) (c) : \(E_{n}=\frac{-13.6}{n^{2}}\left(Z^{2}\right)\)
In first excited state \(E_{\mathrm{H}_{2}}=3.4 \mathrm{eV} \text { and } E_{\mathrm{He}}=-13.6 \mathrm{eV}\)
So, H₂ atom gives excitation energy
(13.6- 3.4 = 10.2 eV) to helium atom
Now, energy of He ion = -13.6 + 10.2 = -3.4 eV
Again, \(E=\frac{-13.6}{n^{2}} \times Z^{2}\)
\(\Rightarrow \quad-3.4=\frac{-13.6}{n^{2}} \times(2)^{2} \Rightarrow n=4\)
(ii) (c): \(\frac{1}{\lambda}=\frac{13.6 Z^{2}}{h c}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)
Here, \(n_{1}=3 \text { and } n_{2}=4 \Rightarrow \lambda=4.8 \times 10^{-7} \mathrm{~m}\)
(iii) (a): \(\text { Kinetic energy, } K \propto \frac{Z^{2}}{n^{2}}\)
\(\frac{K_{\mathrm{H}_{2}}}{K_{\mathrm{He}}}=\left(\frac{Z_{\mathrm{H}_{2}}}{Z_{\mathrm{He}}}\right)^{2}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}\)
(iv) (b): Radius of the permitted orbit is \(r=\frac{n^{2} h^{2} \varepsilon_{0}}{\pi m Z e^{2}}\) For hydrogen atom in ground state, i.e.,|
\(n=1, Z=1 \Rightarrow r=\frac{h^{2} \varepsilon_{0}}{\pi m e^{2}}\)
(v) (a): Angular momentum for hydrogen atom is
\(L=\frac{n h}{2 \pi}\)
For first excited state \(n=2, \quad L=\frac{h}{\pi}\)

(i) (c)
(ii) (d): \(\omega=\frac{v}{r} . \text { Further } v \propto \frac{1}{n} \text { and } r \propto n^{2},\)
\(\text { Hence } \omega \propto\left(1 / n^{3}\right)\)
(iii) (c)
(iv) (b): The energy of n^th Bohr orbit in hydrogen atom is
\(E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV}\)
For lowest orbit, n = 1
\(\therefore \quad E_{1}=-13.6 \mathrm{eV}\)
Thus, the lowest Bohr orbit in hydrogen atom has the least energy
(v) (b)