(i) 6Li3 + 1no \(\rightarrow\)3H1 + 4He2 +Q (energy)
(ii) Q = \(\triangle\)mx931 MeV
Where \(\triangle\)m = 6.01512+1.0086654-4.0026044-3.0100000

a) ²²⁶Ra₈₈-----------------\(\rightarrow\)-²²²Rn₈₆ + ⁴He₂
b) ³²P₁₅------------------ \(\rightarrow\)³²S₁₆ + ⁰e_-1 + \(\gamma \)
c) ³²p¹⁵------------------- \(\rightarrow\)¹¹B₅+⁰e₊₁+  \(\gamma \)

According to the Bohr's theory of hydrogen atom, the angular momentum of revolving electron is given by
\(mvr=\frac { nh }{ 2\pi } \)           ....(i)
where, m = mass of the electron, v = velocity of the electron,
r = radius of the orbit,
h = Planck's constant
n = principal quantum number of the atom.
If an electron of mass m and velocity v is moving in a circular orbit  f radius r, then the centripetal force is given by
\({ F }_{ c }=\frac { { mv }^{ 2 } }{ r } \)          ....(ii)
Also, if the charge on the nucleus is Ze, then the force of electrostatic attraction between the nucleus and the electron will provide the necessary centripetal force.
\(\Rightarrow\)   F_c = F_e
\(\Rightarrow\)   \(\frac { { mv }^{ 2 } }{ r } =\frac { k{ e }^{ 2 } }{ { r }^{ 2 } } \)     [\(\because\) Z=1]
\(\Rightarrow\)    \(r=\frac { { e }^{ 2 }.k }{ { mv }^{ 2 } } \)          ...(iii)
From Eq. (i), we get \(v=\frac { nh }{ 2\pi mr } \)
Putting this value is Eq. (iii). we get
\(r=\frac { k{ e }^{ 2 }.4{ \pi }^{ 2 }{ m }^{ 2 }{ r }^{ 2 } }{ m.{ n }^{ 2 }{ h }^{ 2 } } \)
\(\Rightarrow\) \(r=\frac { { n }^{ 2 }{ h }^{ 2 } }{ k{ e }^{ 2 }.4{ \pi }^{ 2 }m } \Rightarrow r\propto { n }^{ 2 }\)

Energy of electron at n = 2 states is
\(E=\frac { -13.6 }{ { n }^{ 2 } } =\frac { -13.6 }{ \left( { 2 }^{ 2 } \right) } =-3.4ev\)
Now, de-Broglie wavelength of electron is given by
\(\lambda =\frac { h }{ p } =\frac { h }{ \sqrt { 2mk } } \)
\(=\frac { h }{ \sqrt { 2mE } } [\because \left| k \right| =\left| E \right| ]\)
\(=\frac { 6.62\times { 10 }^{ -34 } }{ \sqrt { 2\times 9.1\times { 10 }^{ -31 }\times 3.4\times 1.6\times { 10 }^{ -19 } } } \)
\(=\frac { 6.62\times { 10 }^{ -34 } }{ 10\times { 10 }^{ -25 } } \)
\(=0.662\times { 10 }^{ -9 }\)
\(=6.62\times { 10 }^{ -10 }=6.62\mathring { A } \)

Let N be the number of undecayed nuclei in the sample at time I and t. N nuclei undergo decay in time \(\Delta t\) .
Then \(\frac { -\Delta N }{ \Delta t } \propto N,\frac { -\Delta N }{ \Delta t } =\lambda N\)

where, λ is disintegration constant.
The rate of change in N in time ∆t⇢0, can be expressed as \(\frac { dN }{ N } =-\lambda dt\)
On integrating both sides \(\int _{ { N }_{ 0 } }^{ N }{ \frac { dN }{ N } } =-\int _{ 0 }^{ t }{ \lambda dt } \)
where N₀ is in initial undecayed nuclei.
In \(\frac { N }{ { N }_{ 0 } } =-\lambda t\)
N = N₀e^-λt
mark of N = \(\frac { { N }_{ 0 } }{ 16 } \) in terms of T_1/2 is shown in the figure.

We have
\(R={ R }_{ 0 }{ A }^{ \frac { 1 }{ 3 } }\)
\(\therefore \)  Density \(\rho \) = \(\frac { mA }{ \frac { 4 }{ 3 } \pi \left( { R }_{ 0 }{ A }^{ \frac { 1 }{ 3 } } \right) ^{ 3 } } \)
=\(\frac { m }{ \frac { 4 }{ 3 } \pi { R }_{ 0 }^{ 3 } } \)
Hence is independent of A. (Here m is the mass of the nucleus).

\(2.12\times { 10 }^{ -10 }\ m\ and\ 4.47\times { 10 }^{ -10 }\ m\)

Yes, the nucleons (neutrons and protons) are fundamentals particles. To resolve two objects, say nucleons separated by distance d, the wavelength \(\lambda \) of probing signal must be less than or equal to d. As \(d={ 10 }^{ -15 }m.\) therefore to detect separate parts, if any, inside a nucleon, the electron must have a wavelength ​​​​​​\(\lambda \le { 10 }^{ -15 }m.\) ​Now, \(\lambda =\frac { h }{ p } \ or \ p=\frac { h }{ \lambda } \) and Kinetic energy, 
\(K=pc=\frac { hc }{ \lambda } =\frac { 6.63\times { 10 }^{ -34 }\times 3\times { 10 }^{ 8 } }{ { 10 }^{ -15 } } joule\)
\(K=\frac { 19.89\times { 10 }^{ -11 } }{ 1.6\times { 10 }^{ -19 } } eV\ \simeq { 10 }^{ 9 }eV=1GeV\)               

This is primarily because neutron carries no charge. It is neither attracted nor repelled by nucleus and electrons. Therefore, neutron is best projectile.

This is because only fast moving neutrons of 12 MeV energy can cause fission of \(_{ 92 }{ { U }^{ 238 } }\) nuclei. Such neutrons have less chance of interaction. They escape the fissionable material without causing fission.

Nuclear mass density is of the order of \(PFP-V-II-91/11\). This is \({ 10 }^{ 13 }\) to \({ 10 }^{ 14 }\) times the average atomic mass density, which is of the order of \({ 10 }^{ 3 }\) to \({ 10 }^{ 4\ }kg{ m }^{ -3 }\). Typical mass densities of solids and liquids are of the same order as the atomic mass densities. This is because the atoms are tightly packed in these phases. The typical densities of gases at S.T.P are of the order of \({ 10 }^{ -1 }\ kg{ m }^{ -3 }\)  to  \(1\ kg{ m }^{ -3 }\).

No.of electrons in the nucleus = 0,
No.of protons = atomic number = 11,
No.of (protons+neutrons) = mass number = 24
\(\therefore \) No.of neutrons = 24-11 = 13

Up to quantum number 4, an atom has K, L, M and N shells.In any shell, max.no.of.of electrons is \(2{ n }^{ 2 }\).Therefore,
in K shell, max no.of electrons = \(2{ \left( 1 \right) }^{ 2 }=2\)
in L shell, max no.of electrons = \(2{ \left( 2 \right) }^{ 2 }=8\)
in M shell, max no.of electrons = \(2{ \left( 3 \right) }^{ 2 }=18\)
in N shell, max no.of electrons = \(2{ \left( 4 \right) }^{ 2 }=32\)
Total max.no.of electrons in these shells
= 2 + 8 + 18 + 32 = 60
Hence, number of elements would be 60.

When an electron in hydrogen atom passes from n=4 energy level to n=1 level, max. a number of photons =6, corresponding to transitions \(4\rightarrow 3;3\rightarrow 2;2\rightarrow 1;4\rightarrow 2,3\rightarrow 1,4\rightarrow 1.\)  The minimum number of photons can be one only corresponding to the transition \(4\rightarrow 1.\)