Given, nuclear radius of  ²⁷Al, r₁ = 3.6 fm
Nuclear radius of ¹²⁵Fe, r₂ = ?
A₁ = 27,  A₂ = 125
The nuclear radius is given by 
\(R=R_0 A^{\frac{1}{3}} \Rightarrow R \propto A^{\frac{1}{3}} \)
\(\therefore\) \(\frac{R_2}{R_1}=\left(\frac{A_2}{A_1}\right)^{\frac{1}{3}}=\left(\frac{125}{27}\right)^{\frac{1}{3}}=\frac{5}{3}\)
\(\therefore\)\(R_2=\frac{5}{3} R_1=\frac{5}{3}\)x 3.6 = 6 fm
\(\therefore\) R₂ = 6 fm

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

As, we know that the diode only works in forward biased, so the output is obtained only when positive input is given, so the output waveform is

Resistance of a material can be found out by the slope of the curve V versus I. Part BC of the curve shows the negative resistance as with the increase in current, there is a decrease in voltage

Because the protons are positively charged, so they repel each other. Since, this repulsion force is more, so that an excess of neutrons are required to reduce this repulsion.

\(_{ 6 }{ { C }^{ 11 } }\rightarrow _{ 5 }{ { X }^{ 11 } }+_{ 1 }{ { e }^{ 0 } }(i.e.,\quad { \beta }^{ + })+Q\)
Clearly,  \(_{ 5 }{ { X }^{ 11 } }\) is an isobar of \(_{ 6 }{ { C }^{ 11 } }\).
Mass defect, \(\Delta m\) = 11.011434-(11.009305+0.000545)
= 0.001584 u
Q value of reaction  \(=0.001584\times 931\ MeV\)
= 1.4747 MeV

In case of \(_{ 3 }{ { X }^{ 7 } }\),
\(\frac { neutron\quad number }{ proton\quad number } =\frac { 7-3 }{ 3 } =1.33\)
In case of \(_{ 3 }{ { Y }^{ 4 } }\quad \)
 \(\frac { neutron\quad number }{ proton\quad number } =\frac { 4-3 }{ 3 } =\frac { 1 }{ 3 } =0.33\)
For stability, this ratio has to be close to one.
Obviously, nucleus \(_{ 3 }{ { X }^{ 7 } }\) is more stable than the nucleus \(_{ 3 }{ { Y }^{ 4 } }\).

Both, alpha and beta radiations are affected by both, electric field and magnetic field.

This is because density \(=\frac{m a s s}{\text { volume }}\) and volume of nucleus varies directly as its mass number 

(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) Fission is the process of separating two heavy, unstable atomic nuclei into two lighter nuclei, also releasing energy, but less than with fusion. Fusion is where two light atomic nuclei unite and release energy.

(b) Number of atoms in 100g deuterium, \(\frac{100}{2} \times 6.023 \times 10^{-23}\)
Energy Released- \(\frac{3.27}{2} \times 1.6 \times 10^{-13} \mathrm{~J}\)
Time required t = Total energy(E) / Power of lamp (P)
= 500 \(\times\) 60 \(\times\) 60 \(\times\) 24 \(\times\) 365
t = 1.5768 \(\times\) 10¹⁰ years.

The value of resistance (R) should be increased to keep the reading of ammeter constant, as with the increase in the temperature of a semiconductor, its resistance decreases and current tends to increase.
Or
For graph B,
As the potential barrier in graph B is higher with respect to potential barrier in graph 4, i.e. potential barrier in graph B is increased. So, it is the case of reverse biased condition.
For graph C,
As the potential barrier in graph C is decreased with respect to potential barrier in graph A. So, the graph C is the case of forward biased condition.

The accumulation of negative charges in the p-region and positive charges in the M-region set up a potential difference across the junction. This acts as a barrier and is called potential barrier.

(i) The basic nuclear process involved in the emission of β⁺ during radioactivity is given by
\({ }_{Z}^{A} X \longrightarrow{ }_{Z-1}^{A} Y+\beta^{+}+v\)
(ii) (a) According to question
\({ }_{6}^{11} \mathrm{C} \longrightarrow{ }_{y}^{z} \mathrm{~B}+x+\mathrm{v}\)
For β⁺-decay,
\(z X^{A} \longrightarrow z_{-1} Y^{A}+\beta^{+}+v\)
On comparing Eqs. (i) and (ii), we get
\(y=5, z=11 \text { and } x=\beta^{+}\)
\(\text { (b) }{ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \longrightarrow{ }_{a}^{20} \mathrm{Ne}+{ }_{b}^{c} \mathrm{He}\) 
Helium have 4 mass number and 2 charge number. So reaction will be
\({ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \longrightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_{2}^{4} \mathrm{He}\)
On comparing Eqs. (i) and (ii), we get
a = 10, b = 2 and c = 4

We have the expression for nuclear radius as:
R = R₀A¹/³
Where,
R₀ = Constant.
A = Mass number of the nucleus
Nuclear matter density, \(\rho \ =\frac{Mass\ of\ the\ nucles}{Volume\ of\ the\ nucles}\)
Let m be the average mass of the nucleus.
Hence, mass of the nucleus = mA
\(\therefore \rho=m \frac{A}{\frac{4}{3} \pi R^{3}}=\frac{3 \mathrm{~mA}}{4 \pi\left(R_{0} A^{\frac{1}{3}}\right)^{3}}=\frac{3 m A}{4 \pi R_{0}^{3} A}=\frac{3 \mathrm{~m}}{4 \pi R_{0}^{3}}\)
Hence, the nuclear matter density is independent of A. It is nearly constant.

(i) According to Bohr's principle, electrons revolve in a stationary orbit of which energy and momentum are fixed. The momentum of electrons in the fixed orbit is given by \(\frac { nh }{ 2\pi } \) (where n = the number of orbits). According to de-Broglie's hypothesis, the electron is associated with wave character. Hence, a circular orbit can be taken to be a stationary energy state only. if it contains an integral number of de-Broglie wavelengths, i.e., 2\(\pi\)r = n\(\lambda\)
(ii) According to question,

\({ E }_{ B }-{ E }_{ C }=\frac { hc }{ { \lambda }_{ 1 } } \quad \quad ...(i)\)
\({ E }_{ A }-{ E }_{ B }=\frac { hc }{ { \lambda }_{ 2 } } \quad \quad ...(ii)\)
\({ E }_{ C }-{ E }_{ A }=\frac { -hc }{ { \lambda }_{ 3 } } \quad ...(iii)\)
On adding Eqs. (i), (ii) and (iii), we get
E_B - E_C + E_A - E_B + E_C - E_A
\(=hc\left( \frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } -\frac { 1 }{ { \lambda }_{ 3 } } \right) \)
\(\frac { 1 }{ { \lambda }_{ 3 } } =\frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } \Rightarrow { \lambda }_{ 3 }=\frac { { { \lambda } }_{ 1 }{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } } \)

(i) (c) The energy required to free the weakly bound electron from the dopant is about 0.7 eV.
(ii) (d) Here,
Number of intrinsic charge carriers,
\(n_i=2.0 \times 10^{10} \mathrm{~cm}^{-3}\)
Number of holes, n_h = 8 \(\times\) 10³ cm^-3
We know that,
\(\begin{aligned} n_e n_h & =n_i^2 \end{aligned}\)
\(\Rightarrow\)  \(\begin{aligned} n_e & =\frac{n_i^2}{n_h}=\frac{\left(2.0 \times 10^{10}\right)^2}{8 \times 10^3} \\ \end{aligned}\)
\(\begin{aligned} =\frac{4.0 \times 10^{20}}{8 \times 10^3}=0.5 \times 10^{17} \mathrm{~cm}^{-3} \end{aligned}\)
= 5 \(\times\) 10¹⁶ cm^-3
n_e = 5 \(\times\)10⁺²² m^-3
(iii) (c) In the p-section, holes are the majority carriers; while in n-section,the majority carriers are electrons. Due to the high concentration of different types of charge carriers in the two section, holes from p-region diffuse into n-region and electrons from n-region diffuse into p-region.
Or
(iii) (d) Initially, during the formation of p-n junction, diffusion of majority charge carriers take place from p-side to n-side which constitute the diffusion current. Thus, diffusion current is large and the drift current is small.
(iv) (d) Given,
AC voltage V = 0.5 sin (100 \(\pi\)t) V
Here, V₀ = 0.5
\(\begin{aligned} \omega & =100 \pi \end{aligned}\)
\(\begin{aligned} 2 \pi f & =100 \pi \end{aligned}\)
\(\Rightarrow\) f = 50 Hz
The frequency of the output voltage across half- wave rectifier is 50Hz and across full wave rectifier is 100Hz.

(i) (a) : Let the number of fissions per second be n. Energy released per second
\(=n \times 200 \mathrm{MeV}=n \times 200 \times 1.6 \times 10^{-13} \mathrm{~J}\)
Energy required per second = power x time
\(=1 \mathrm{~kW} \times 1 \mathrm{~s}=1000 \mathrm{~J}\)
\(\therefore \quad n \times 200 \times 1.6 \times 10^{-13}=1000\)
\(\text { or } \quad n=\frac{1000}{3.2 \times 10^{-11}}=\frac{10}{3.2} \times 10^{13}=3.125 \times 10^{13}\)
(ii) (a)
(iii) (b): As only 0.1% of the original mass is converted into energy, hence out of 1 kg mass 1 g is converted into energy.
\(\therefore\) Energy released during fission, \(E=\Delta m c^{2}\)
\(=1 \mathrm{~g} \times\left(3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}\right)^{2}=10^{-3} \times 9 \times 10^{16} \mathrm{~J}=9 \times 10^{13} \mathrm{~J}\)
(iv) (a)
(v) (a)