Distance of closest approach is the distance between the centre of nucleus and the point from which the alpha particle approaching directly to the nucleus returns.
Impact parameter is the perpendicular distance of the velocity vector of the alpha particle from the central line of the nucleus, when the particle is far away from the atom.

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.

\(N=\frac { { N }_{ 0 } }{ { 16 }^{ ' } } \)
Where 30 years
\(N={ N }_{ 0 }\left( \frac { 1 }{ 2 } \right) ^{ n }\)
\(\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 2 } \right) ^{ 4 }\)
No.of half lives = 4
\(4=\frac { Time \ of \ disintegration }{ half \ life \ period } \)
\(\Rightarrow \frac { 30 \ years }{ 4 } =half \ life \ period\)
Half-life period = 7.5 years

For second excited state, n = 3
Hence two possible transition of the Lyman series: 3→1 and 2→1
Wavelength for transition 3→1, n_f = 1, n_i = 3
\(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { n }_{ f }^{ 2 } } -\frac { 1 }{ { n }_{ i }^{ 2 } } \right) \)
\(\frac { 1 }{ \lambda } =1.1\times { 10 }^{ 7 }\left( \frac { 1 }{ 1 } -\frac { 1 }{ 9 } \right) \)
\(=1.1\times { 10 }^{ 7 }\left( \frac { 8 }{ 9 } \right) \)
⇒ λ = \(\frac { 9 }{ 8\times 1.1\times { 10 }^{ -7 } } \)
= 1.023 x 10^-7
= 102. 3nm
For transition 2→1, n_f =1, n_i=2
\(\frac { 1 }{ \lambda } =1.1\times { 10 }^{ 7 }\left( 1-\frac { 1 }{ 4 } \right) \)
⇒ λ = 212 nm

\((i) \ -1.7eV; \ 0.85 \ eV; \ (ii) \ 3\times { 10 }^{ 15 }Hz\)
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, \(\Delta E={ E }_{ 4 }-{ E }_{ 1 }\)
\(hv=-0.85-(-13.6)eV=12.75 \ eV\)
\(v=\frac { 12.75\times 1.6\times { 10 }^{ -19 } }{ 6.6\times { 10 }^{ -34 } } =3\times { 10 }^{ 15 }Hz\)

\(0.53\mathring { A } ;2.19\times { 10 }^{ 6 }{ ms }^{ -1 };6.6\times { 10 }^{ 15 }Hz\)

\(Here,\ T=20s,\ \lambda =? \ t=?\frac { N }{ { N }_{ 0 } } =1-\frac { 7 }{ 8 } =\frac { 1 }{ 8 } \)
\(\lambda =\frac { 0.693 }{ T } =\frac { 0.693 }{ 20 } =0.0346{ s }^{ -1 }\)
\(As \ \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 1 }{ 2 } \right) }^{ n }=\frac { 1 }{ 8 } \)
\(n=3, \ t=3 \ T=3\times 20s=60s\)

(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
\(=hc\left( \frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } -\frac { 1 }{ { \lambda }_{ 3 } } \right) \)
\(\frac { 1 }{ { \lambda }_{ 3 } } =\frac { 1 }{ { \lambda }_{ 1 } } +\frac { 1 }{ { \lambda }_{ 2 } } \Rightarrow { \lambda }_{ 3 }=\frac { { { \lambda } }_{ 1 }{ \lambda }_{ 2 } }{ { \lambda }_{ 1 }+{ \lambda }_{ 2 } } \)

Let a sample of radioactive material have N₀ nuclei, at t = 0
At time t, Number of nuclei = N
As per the decay law
\(-\frac{dN}{dt}=\lambda N\)
\(\Rightarrow \ \int _{ { N }_{ 0 } }^{ N }{ \frac { dN }{ N } } =\int _{ 0 }^{ t }{ -\lambda } dt\)
\(\Rightarrow \ { \left( \log _{ e }{ N } \right) }_{ { N }_{ 0 } }^{ N }=-\left( { t }_{ 0 } \right) { t }^{ 1/2 }\)
\(\frac{N}{N_0}=e^-\lambda^ t\)
\(\Rightarrow \ \ N={ N }_{ 0 }{ e }^{ -\lambda t }\)
After one half life, Number of nuclei becomes  \(\frac{N_0}{2}\)
\(\Rightarrow \ \frac { N_{ 0 } }{ 2 } =N_{ 0 }e^{ - }\lambda ^{ T }1/2\)
\(\Rightarrow\)  2 = \(e^\lambda \)^T1/2
\(\Rightarrow\)  log_e 2 = \({ \lambda T }_{ \frac { 1 }{ 2 } }\)
\(\Rightarrow\)  \({ \lambda T }_{ \frac { 1 }{ 2 } }\)  = 0.6931
\(\Rightarrow\)  \({ T }_{ \frac { 1 }{ 2 } }\) = \(\frac{0.6931}{\lambda}\)

\(In \ _{ 26 }{ { Fe }^{ 56 } },\ no.of \ protons \ 26;\)
no. of neutrons = 56 - 26 = 30
\( \therefore \ Mass\ defect=26{ m }_{ p }+30{ m }_{ n }-{ M }_{ Fe }\)
\(=26\times 1.007825+30\times 1.008665-55.934939\)
\( =26.20345+30.25995-55.934939\)
\( =0.528461\ u\)
\( B.E/nucleon=\frac { 0.528461\times 931 }{ 56 } =8.79MeV/N\)

\(mvr = \frac {nh}{2\pi}\)
\(\frac { { mv }^{ 2 } }{ r } =\frac { 1 }{ 4\pi { \varepsilon }_{ 0 } } \frac { { e }^{ 2 } }{ { r }^{ 2 } } \)
\(r=\frac { { e }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }{ mv }^{ 2 } } \)
\(r=\frac { { Ze }^{ 2 } }{ { 4\pi \varepsilon }_{ 0 }m{ \left( \frac { nh }{ 2\pi mr } \right) }^{ 2 } } \)
\(\Rightarrow\)  \(r=\frac { { \epsilon }_{ 0 }{ n }^{ 2 }{ h }^{ 2 } }{ { \pi me }^{ 2 } } \)
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
\(\frac { 1 }{ \lambda } =R\left( \frac { 1 }{ { 2 }^{ 2 } } -\frac { 1 }{ { 3 }^{ 2 } } \right) \)
\(\lambda =\frac { 36 }{ 5R } \)