Energy produced per second in reactor = 1 W = 1 J/s
Energy produced per fission = 200 MeV = 200 x 10⁶ x 1.6 x 10^-19 J
= 3.2 x 10¹¹ J
Number of fissions per second required \(=\frac{\text { Energy produced per second in reactor }}{\text { Energy produced per fission }} \)
\(=\frac{1}{3.2 \times 10^{11}}=\frac{10 \times 10^{10}}{3.2}=3.125 \times 10^{10} \)
Number of fissions per second = 3.125 x 10¹⁰ 

Kinetic energies,
KE_Ai = 0
KE_Af = KE_Bt
It should obey law of conservation of energy
Total Initial KE = Total final KE
\(\mathrm{KE}_{\mathrm{A}_{i}}+\mathrm{KE}_{\mathrm{B}_{i}}=\mathrm{KE}_{\mathrm{A}_{\mathrm{t}}}+\mathrm{KE}_{\mathrm{B}_{\mathrm{t}}} \)
\(0+\mathrm{KE}_{\mathrm{B}_{i}}=\mathrm{KE}_{\mathrm{A}_{\mathrm{f}}}+\mathrm{KE}_{\mathrm{A}_{t}} \)
\(\mathrm{KE}_{\mathrm{B}_{\mathrm{i}}}=2 \mathrm{KE}_{\mathrm{A}_{\mathrm{f}}} \)
Minimum energy required for excite the hydrogen atom is 10.2eV
Hence minimum Kinetic energy of H_B is
\(\mathrm{KE}_{\mathrm{B}_{1}}=2 \times 10.2 \mathrm{eV}=20.4 \mathrm{eV}\)

\(\mathrm{T}_{\frac{1}{2}}=5.01 \text { days} \), \(\mathrm{R}_{o}=1 μ \mathrm{Ci}=3.7 \times 10^{10} \times 10^{-6} \)  decays per second
\(if\ \mathrm{t}=1\ year\ \mathrm{R}= ? \), \(\tau=? \), \(\lambda=? \), \(\mathrm{~N}_{o}=? \)
a) \(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\lambda t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{\mathrm{~T}_{1 / 2} }t} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-\frac{0.6931}{5.01} \times 365} \)
\(\mathrm{R}=\mathrm{R}_{o} \mathrm{e}^{-50.5}=\mathrm{R}_{o} \times 1.17 \times 10^{-22} \)
\(\mathrm{R}=1 \mu \mathrm{Ci} \times 1.17 \times 10^{-22}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
\(\mathrm{R}=1.17 \times 10^{-22} \mu \mathrm{Ci} \)
b) \(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01}=0.1383 \text { day }^{-1} \)
\(\lambda=\frac{0.6931}{T_{\frac{1}{2}}}=\frac{0.6931}{5.01 \times 24 \times 60 \times 60}=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
\(\lambda=1.6 \times 10^{-6} \mathrm{~s}^{-1} \)
c) \(\tau=\frac{1}{\lambda}=\frac{1}{0.1383}=7.23 \text { days } \)
\(\tau=7.23 \text { days } \)
\(R_{o}=\lambda N_{o} \)
\(N_{o}=\frac{R_{o}}{\lambda}=\frac{3.7 \times 10^{10} \times 10^{-6}}{1.6 \times 10^{-6}}=2.3 \times 10^{10} \)
\(N_{o}=2.3 \times 10^{10} \)

A = 108, Z  =47, N = 108 - 47 = 61
m_p = 1.007825 u, m_n = 1.008665 u, M = 107.905949 u
(a) \(\Delta \mathrm{m}=Z \mathrm{~m}_{\mathrm{P}}+\mathrm{Nm}_{\mathrm{n}}-\mathrm{M} \)
\(\Delta \mathrm{m}\) = (47 x 1 .007825 + 61 x 1 .008665 - 107 .905949)
\(\Delta \mathrm{m}\) = 47.367775 + 61.528565 -107.905949
\(\Delta \mathrm{m}\) = 108.89634 - 107.905949
\(\Delta \mathrm{m}\) = 0.990391 u
\(\mathrm{BE}=\Delta \mathrm{m} \times 931 \mathrm{MeV} \)
BE = 0.990391 x 931 MeV = 922.054 MeV
(c) \(\overline{\mathbf{B E}}=\frac{\mathbf{B E}}{\mathbf{A}} \)
\(\overline{\mathrm{BE}}=\frac{922.054}{108}=8.537 \mathrm{MeV}=8.5 MeV\)

Wavelength of incident radiation = 97.5 nm = 97.5 x 10^-9 m
Energy of hydrogen atom in its ground state = -13.6 eV
(a) Principal quantum number n = ?
(b) (i) Number of possible transitions = ?
(ii) Total number possible lines = ?
(a) Energy absorbed by Hydrogen atom
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{97.5 \times 10^{-9}} \mathrm{~J} \)
\(E=\frac{h c}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{97.5 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV} \)
E = 12.74 eV
Energy of the electron in first orbit of Hydrogen is -13.6 ev
E_n = -13.6 + 12.74 = -0.86 eV
We know that
\(\mathrm{E}_{\mathrm{n}} =-\frac{13.6}{\mathrm{n}^{2}} \)
\(-0.86 =-\frac{13.6}{\mathrm{n}^{2}} \)
\(\mathrm{n}^{2} =15.88 \)
\(\mathrm{n} \cong 4 \)
(b) (i) By using arithmetic progression, For the principle quantum number "n",
Total number of possible transition form level n is \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)
(ii) Total number of possible transitions form level 4 is 3
Total number of possible transitions form level 3 is 2
Total number of possible transition form level 2 is 1
Hence total number of possible transitions is 3 + 2 + 1 = 6