In a nuclear reaction, the sum of the masses of the target nucleus \((_{1}^{2}{H})\) and the bombarding particle \((_{1}^{2}{H})\) may be greater less than the sum of the masses of the product nucleus \((_{1}^{3}{He})\) and the outgoing particle \((_{0}^{1}{n})\). So, from the law of conservation of mass-energy, some energy (3.27 MeV) is evolved or involved iRjjl nuclear reaction. This energy is called Q -value of the nuclear reaction.
Density of nuclear matter is the ratio of mass of the nucleus and its volume.
Density of the nuclear matter
= \({{Mass\ of \ nucleus}\over{Volume\ of\ nucleus}}\)               ..(i)
If m is average mass of a nucleon and R is the nuclear radius, then mass of nucleus = mA, where, A is the mass number of the element.
Volume of the nucleus = \({{4}\over{3}}\pi{R}^{3}\)
= \({{4}\over{3}}(\pi{R}_{0}{A}^{1/3})^{1/3}={{4}\over{3}}\pi{R}^{3}_{0}A\)
But put the value in Eq. (i).
Thus, density of nucleus = \({ { mA } \over { { { 4 } \over { 3 } } }\pi{R}^{3}_{0} A}={ { 3m } \over { 4\pi{R}^{3}_{0} } }\)
where, m = mass of one nucleon
A = mass
As, m and Ro are constants, therefore density of the nuclear matter is the same for all elements. Now, using m = 1.66 x 10^-27 kg.
\(={{3\times1.66\times{10}^{-27}}\over{4\pi{R}^{3}_{0}}}\)
Using Ro = 1.1 x 10^-15 m
and density = 2.97 x 10¹⁷ kg m^-3
which shows that the density is independent of mass number A.

Stopping potential is more for the curve corresponding to frequency V₁ 
Stopping po ential is directly proportional to the frequency of incident radiation.

Magnetic momentum,
\(\mu=\left(\frac{e}{2m}\right)L\)
where (-) indicates \(\mu\) direction to L.
As  Bohr's atomic model
L = mvr
\(\mu = - \left(\frac{e}{2m}\right)\times mvr\)
\(\mu = \frac {evr}{2}\)
Energy levels of hydrogen atom
\({E}_{n}=\frac{{2\pi}^{2}{m}{K}^{2}{e}^{4}}{{n}^{2}{h}^{2}}\)
For lowest energy level n = 1
\({E}_{n}=-\frac{{2\pi}^{2}{m}{K}^{2}{e}^{4}}{{n}^{2}{h}^{2}}\)
\({E}_{2}=-\frac{{2\pi}^{2}{m}{K}^{2}{e}^{4}}{{h}^{2}}\)
\(E_1=-\left(\frac{{4\pi}^{2}{m}{Ke}^{2}}{{h}^{2}}\right)\frac{{Ke}^{2}}{2}\)
While   \(r=\frac { { h }^{ 2 } }{ { 4n }^{ 2 }{ kme }^{ 2 } } \)
\({ E }_{ n }=-\frac { { Ke }^{ 2 } }{ 2r } =\frac { -3.6 }{ { n }^{ 2 } } eV\)
\(\therefore \mu =\frac { eVr }{ 2 } \)
\({ E }_{ 1 }=-\frac { Kev.er }{ { 2vr }^{ 2 } } =-13.6\)
\(\frac { k\mu e }{ { vr }^{ 2 } } =-13.6\)
 \(\mu =13.6\left( \frac { { ur }^{ 2 } }{ Ke } \right) \)

In the given nuclear reaction,
Mass of reactants = \(m(_{ 1 }{ { H }^{ 2 } })+m(_{ 1 }{ { H }^{ 3 } })\)
\( =2.014102+3.016049\)
\( =5.030151u\)
Mass of products
\(m(_{ 2 }{ { He }^{ 4 } })+m(_{ 0 }{ { n }^{ 1 } })\)
\(=4.002603+1.008665\)
\( =5.011268u\)
\(Mass \ defect,\Delta m=5.030151-5.011268\)
\(=0.018883u\)
\(Energy \ released=0.018883\times 931 \ MeV\)
\( =17.58MeV\)

0.9 eV

Einstein's photoelectric equation,
K.E. of photoelectron = Incident energy of photons - Work function
or  K.E = hv - W₀
or  K.E = hv - hv₀
where v₀ is called threshold frequency
Threshold Frequency : For a given metal, there exists a certain minimum frequency of the incident radiation below which no emission of photoelectrons takes place. This frequency is called threshold frequency.

It is the minimum negative potential given to the anode in a photocell for which the photoelectric current becomes zero. If \({ v }_{ 0 }\) is the stopping potential, then maximum K.E. of emitted photoelectron is
\({ \left( K.E \right) }_{ max }={ eV }_{ 0 }=hv-{ \phi }_{ 0 }\)
\(V_{ 0 }=\frac { hv }{ e } -\frac { { \phi }_{ 0 } }{ e } \)
Threshold frequency  It is the minimum frequency of the incident radiation for which just emission of photoelectrons takes place from a metal surface without any K.E. If \(V_{ 0 }\)is the threshold frequency, then using Einstein's photoelectric equation
\(0={ hv }_{ 0 }-{ \phi }_{ 0 }\quad or\quad V_{ 0 }=\frac { { \phi }_{ 0 } }{ h } \)

The mobility of electrons in \(n\)-type semiconductor is more than the mobility of holes in \(p\)-type semiconductor. Since the \(n\)-type semiconductor has electrons as majority carriers and holes as minority carriers, whereas the \(p\)-type semiconductor has holes as majority carriers and electrons as minority carriers, therefore mobility of  \(n\)-type is more than that of  \(p\)-type.

The correct statement is (c).
When a forward bias is applied to a p-n junction, it lowers the value of potential barrier. In the case of a forward bias, the potential barrier opposes the applied voltage. Hence, the potential barrier across the junction gets reduced.

(a) (i) Electron moving in a circular orbit around the nucleus would get accelerated, therefore it would spiral into the nucleus, as it looses its energy.
(ii) It must emit a continuous spectrum. According to Bohr's model of hydrogen atom
(i) Electron in an atom can revolve in certain stable orbits without the emission of radiant energy

The two processes are
(i) Diffusion 
(ii) Drift
Diffusion : Holes diffuse from p-side to n-side (p \(\rightarrow \) n) and electrons diffuse from n-side to p-side (n\(\rightarrow \) p)
Drift: The motion of charge carriers, due to the applied electric field \((\vec { E } )\)  which results in drifting of holes along E and of electrons opposite to that of electric field \((\vec { E } )\)).
(b) Name of device: Zener Diode

Working:
Any increase I decrease in the input voltage results in an increase I decrease of the voltage drop across R,; without any change in voltage across the Zener diode. Thus Zener diode acts as voltage regulator.

Reverse biasing

The VI characteristics are obtained by connecting the battery, to the diode, through a potentiometer (or rheostat). The applied voltage to the diode is changed. The values of current, for different values of voltage, are noted and a graph between V and I is plotted. The V-I characteristics, of a diode, have the form shown here.

(b) The circuit diagram, for the photodiode, is shown here.

The photodiode is illuminated by optical signal, whose photon energy is greater than the energy gap of the semiconductor used.
The electric field, at the junction, separates the electrons and holes and thus gives rise to an emf.
When an external load is connected, a (photo) current flows through it. The magnitude of this current is proportional to the intensity of light incident on the photodiode.

(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.

\(mvr = \frac {nh}{2\pi}\)
\(\frac { { mv }^{ 2 } }{ r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { e }^{ 2 } }{ { r }^{ 2 } } \)
\(r=\frac { { e }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }{ mv }^{ 2 } } \)
\(r=\frac { { Ze }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }m{ \left( \frac { nh }{ 2\pi mr } \right) }^{ 2 } } \)
\(\Rightarrow\)  \(r=\frac { { \epsilon }_{ 0 }{ n }^{ 2 }{ h }^{ 2 } }{ { \pi me }^{ 2 } } \)
Potential energy U \(=-\frac { 1 }{ 4{ \pi \epsilon }_{ 0 } } .\frac { { e }^{ 2 } }{ { r } } \)
\(=\frac { { me }^{ 4 } }{ { 4\epsilon }_{ 0 }{ n }^{ 2 }{ h }^{ 2 } } \) 
\(KE=\frac { 1 }{ 2 } { mv }^{ 2 }=\frac { 1 }{ 2 } m{ \left( \frac { nh }{ 2\pi mr } \right) }^{ 2 }\)
\(=\frac { { n }^{ 2 }{ h }^{ 2 }{ \pi }^{ 2 }{ m }^{ 2 }{ e }^{ 4 } }{ { 8\pi }^{ 2 }{ me }_{ 0 }^{ 2 }{ n }^{ 2 }{ h }^{ 2 } } \)
\(KE=\frac { { me }^{ 4 } }{ { 8\varepsilon }_{ 0 }^{ 2 }{ n }^{ 2 }{ h }^{ 2 } } \)
TE = KE + PE
\(=-\frac { { me }^{ 4 } }{ { 8\epsilon }_{ 0 }^{ 2 }{ n }^{ 2 }{ h }^{ 2 } } \)
(b)Rydberg formula: For first member of Lyman series
  \(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { 1 }^{ 2 } } -\frac { 1 }{ { 2 }^{ 2 } } \right) \)
\(=\frac { 4 }{ 3R } \)
For first member of Balmer Series
\(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { 2 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 2 } } \right) \)
\(\lambda =\frac { 36 }{ 5R } \)

(i) Davisson and Germer observed diffraction patterns of slow moving electrons. And G.P. Thomson observed a diffraction pattern of fast moving electrons. As diffraction is essentially a wave phenomenon, therefore, it was concluded that wave must be associated with moving electrons.
(ii) (a) At rest, v = 0, \(=\frac { h }{ mv } =\frac { h }{ m\times 0 } =\infty \)
(b) Particle moving with infinite speed, v = \(\infty \) 
\(\lambda =\frac { h }{ mv } =\frac { h }{ m\times \infty } \)
(iii) The dual nature of moving material particles reveals in a way the nature of Almighty God. He is in a visible form (i.e., Sakar) like a visible particle and also without any form (i.e., Nirakar) like a wave.It depends on us how we realize him. ​