\(2.27\times { 10 }^{ -14 }m\)
Here, \(b=?\ KE=5MeV=5\times 1.6\times { 10 }^{ -13 }J\)
\(\theta =90°,\ \quad Z=79\)
\( b=\frac { Z{ e }^{ 2 }cot{ \theta }/{ 2 } }{ 4\pi { \epsilon }_{ 0 }(KE) } \)
\( =\frac { 9\times { 10 }^{ 9 }\times 79{ \left( 1.6\times { 10 }^{ -19 } \right) }^{ 2 }cot45° }{ 5\times 1.6\times { 10 }^{ -13 } }\)
\( b=2.27\times { 10 }^{ -14 }m\)

(i) n-type
(ii) \(4\times 10^{ 4 }/m^{ 3 }\) 
(iii) Energy gap decreases with doping
Here,
\({ n }_{ i }=6\times 10^{ 8 }m^{ -3 };{ n }_{ e }=9\times 10^{ 12 }m^{ -3 }\)
\( { n }_{ h }=\frac { { n }_{ i }^{ 2 } }{ { n }_{ e } } =\frac { \left( 6\times 10^{ 8 } \right) ^{ 2 } }{ 9\times 10^{ 12 } } =4\times 10^{ 4 }m^{ -3 }\)
As, after doping, \({ n }_{ e }>{ n }_{ h }\) so the new semiconductors is n-type. Energy gap decreases with doping.

Total B.E of \(_{ 6 }{ { C }^{ 12 } }\)\(​​​​​​=12\times 7.68=92.16\quad MeV\)
Total B.E of \(_{ 6 }{ { C }^{ 13 } }\)  \(=13\times 7.47=97.11\quad MeV\)
As \(_{ 6 }{ { C }^{ 13 } }\) has one excess neutron than \(_{ 6 }{ { C }^{ 12 } }\),
\(\therefore \) Energy required to remove a neutron
= 97.11-92.16 = 4.95 MeV

\(In \ _{ 15 }{ { P }^{ 31 }, }number \ of \ protons=15\)
number of neutrons = 31−15 = 16
Mass defect,
\( \Delta m=15\times 1.00782+16\times 1.00865-30.97376\)
\(=0.28194amu\)
\(B.E./nucleon=\frac { 0.28194\times 931 }{ 31 } =8.47MeV/N\)

10.36V

196.43 V

Given, forward biased resistance = \(25\Omega \)
Reverse biased resistance = \(\infty \)
As the diode in branch CD is in reverse biased which having resistance infinite.
so  \(I_{ 3 }\)= 0
Resistance in branch AB = 25 + 125 = 150\(\Omega \)(say R₁)
Resistance in branch EF = 25 + 125 = 150\(\Omega \)(say R₂)
AB is parallel to EF.
so, resultant resistance, \(\frac { 1 }{ { R }^{ \prime } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 1 }{ 150 } +\frac { 1 }{ 150 } =\frac { 2 }{ 150 } \)
\(\Rightarrow { R }^{ \prime }=75\Omega \)
Total resistance, \(R={ R }^{ \prime }+25=75+25=100\Omega \)
Current, \({ I }_{ 1 }=\frac { V }{ R } =\frac { 5 }{ 100 } =0.05A\)
\({ I }_{ 1 }={ I }_{ 4 }+{ I }_{ 2 }+{ I }_{ 3 }\quad [Here,\quad { I }_{ 3 }=0]\)
\(\\ { I }_{ 1 }={ I }_{ 4 }+{ I }_{ 2 }\)
Here, the resistances R₁ and R₂ are same
\({ I }_{ 4 }={ I }_{ 2 }\)
\(\\ { I }_{ 1 }=2{ I }_{ 2 }\)
\(\\ \Rightarrow { I }_{ 2 }=\frac { { I }_{ 1 } }{ 2 } =\frac { 0.05 }{ 2 } =0.025A\)
\(\\ { I }_{ 4 }=0.025A\)
\(\\ { I }_{ 1 }=0.05A\)
\(\\ { I }_{ 2 }=0.025A\)
\(\\ { I }_{ 3 }=0\)
\(\\ { I }_{ 4 }=0.025A\) 

(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

(i) \(\alpha \) rays
(ii) \(\gamma \)  rays
(iii) \(\beta \)  rays

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,
\(r_{r b}=10 \mathrm{~V} / 1 \mu \mathrm{A}=1.0 \times 10^{7} \ \Omega\) 

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, wavelength of light,
A = 6000\(\overset{o}{A}\) = 6000 x 10^-10 m
\(\therefore\) Energy of the light photon,
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6000 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}=2.06 \mathrm{eV}\)
The incident radiation which is detected by the photodiode having energy should be greater than the band gap. So, it is only valid for diode D₂ . Then, diode D₂ will detect this radiation.

Given, input frequency = 50 Hz
For a half-wave rectifier, the output frequency is equal to the input frequency.
\(\therefore\) Output frequency = 50 Hz
For a full-wave rectifier, the output frequency is twice the input frequency.
\(\therefore\)  Output frequency = 2 x 50 = 100 Hz.

(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

In the given circuit, diode D₁ is forward biased with forward resistance of 50\(\Omega\) and D₂ is reversed biased, which offers infinite backward resistance.
\(\therefore \ I=\frac{\text { Applied voltage }}{\text { Total resistance }}\)
\(I=\frac{6}{50+150+100}\)
\(I=\frac{6}{300}=0.02 \mathrm{~A}\)
Therefore, current through 100 \(\Omega\) resistance = 0.02 A

From the given circuit diagram, diode D₁, is in forward bias whereas diode D₂ is in reverse bias.
Hence, no current will flow through diode D₂.
\(\therefore\)    I₂ = 0
Now, circuit diagram is reduced as

\(\therefore\) Equivalent resistance, R = 3 + 3 = 6\(\Omega\)
Now, Current \(I_1=\frac{V}{R}=\frac{6}{6}=1 \mathrm{~A}\)

The given nuclear reaction is:
\({ }_{1}^{1} H+{ }_{1}^{3} H \rightarrow_{1}^{2} H+{ }_{1}^{2} H\)
It is given that:
Atomic mass \(m\left({ }_{1}^{1} H\right)=1.007825 \mathrm{u}\)
Atomic mass \(m\left({ }_{1}^{3} H\right)=3.016049 \mathrm{u}\)
Atomic mass \(m\left({ }_{1}^{2} H\right)=2.014102 u\)
According to the question, the Q-value of the reaction can be written as:
\(Q=\left[m\left({ }_{1}^{1} H\right)+m\left({ }_{1}^{3} H\right)-2 m\left({ }_{1}^{2} H\right)\right] c^{2}\)
\(=[1.007825+3.016049-2 \times 2.014102] c^{2}\)
\(Q=-0.00433 \times 931.5=-4.0334 \mathrm{MeV}\)
The negativeQ-value of the reaction shows that the reaction is endothermic.
The given nuclear reaction is:
\(\text { (ii) }{ }_{6}^{12} \mathrm{C}+{ }_{6}^{12} \mathrm{C} \rightarrow{ }_{10}^{20} \mathrm{Ne}+{ }_{2}^{4} \mathrm{He}\)
t is given that:
Atomic mass of \(m\left({ }_{6}^{12} C\right)=12.0 u\)
Atomic mass of \(m\left(\begin{array}{l} 20 \\ 10 \end{array}\right)=19.992439\)
Atomic mass of \(m\left({ }_{2}^{4} \mathrm{He}\right)=4.002603 \mathrm{u}\)
The Q-value of this reaction is given as:
\(Q=\left[2 m\left({ }_{6}^{12} C\right)-m\left({ }_{10}^{20} N e\right)-m\left({ }_{2}^{4} H e\right)\right] c^{2}\)
\(=[2 \times 12.0-19.992439-4.002603] c^{2}\)
\(=\left(0.004958 c^{2}\right) u\)
\(=0.004958 \times 931.5=4.618377 \mathrm{MeV}\)
The positive Q-value of the reaction shows that the reaction is exothermic.

From the given situation, nuclear reaction is given as
\(P^{240} \longrightarrow Q^{10}+R^{130}+\Delta E\)
Energy released in the fission
\(\Delta E=110 \times 8.4+130 \times 8.5-240 \times 7.6\)
= 924 + 1105 - 1824
= 205 MeV