One electron volt is defined as the amount of energy acquired by an electron when it is accelerated under a potential difference of 1 volt.

According to mass energy equivalence established by Einstein E = mc². If B represents binding energy of hydrogen atom (= 13.6 eV), the equivalent mass of this energy \(=B / c^2\)
Hence, mass of a H-atom \(=m_p+m_e-B / c^2\)
It is less than sum of the masses of a proton and an electron.

Plot the graph between and potential energy of a pair of nucleons as a function of their separation

(i) For distance less than 0.8 fm, negative PE decreases to zero and then becomes positive 
(ii) for distances larger than 0.8 fm, negative PE goes on decreasing.

We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of 10⁷⁰h. This leads to a very high value of quantum levels n of the order of 10⁷⁰. For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

From the given plot, we can conclude that, a very heavy nucleus A = 240 has lower E_bn compared to that of a nucleus with A = 120.Thus, if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. Energy would be released in this process which is known as nuclear fission.

Also, when two light nuclei (A ≤ 10)join to form a heavier nucleus, E_bn of fused heavier nuclei is more than the E_bn of lighter nuclei. Energy would be released in this process, which is known as nuclear fusion.

(i) Nuclear fusion will occur when the kinetic energy of colliding nuclei is enough to overcome the strong electrostatic forces of repulsion between the protons. For this, high temperature is required.
(ii) The density of nuclei should also be very high to increase the number of collisions.

In an ordinary atom, as a first approximation, we ignore the motion of the nucleus, being too heavy. In a positronium atom, a positron replaces proton of hydrogen atom. As electron and positron masses are equal, the motion of the positron cannot be ignored. We consider motion of electron and positron about their centre of mass. A detailed analysis (beyond the scope of this book) shows that formulae of Bohr model apply to positronium atom provided that we replace \({ m }_{ e }\) by what is known as reduced mass of the electron. For positronium, the reduced mass is \({ m }_{ e }/2.\) In the transition n = 2 to n = 1, the wavelength of radiation emitted is double than that of the corresponding radiation emitted for a similar transition in hydrogen atom, which has a wavelength of \(1217\mathring { A } \) ; and hence is equal to \(2\times 1217=2434\mathring { A } .\) This radiation lies in the ultra-violet part of the electromagnetic spectrum.

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

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
\(\Rightarrow\)  \({ T }_{ \frac { 1 }{ 2 } }\) = \(\frac{0.6931}{\lambda}\)

(i) An electron undergoes transition from second excited state to the first excited state which corresponds to Balmer series and then to the ground state which corresponds to Lyman series.
(ii) The wavelength of the emitted radiations in the two cases.

We know that,\(\lambda=\frac{h c}{\Delta E}\)
Frorn n₃ \(\rightarrow\) n2,
\(\lambda_{1}=\frac{h c}{E_{3}-E_{2}}\)
\(=\frac{h c}{(-1.5)-(-3.4)}=\frac{h c}{1.9}\)
From n2 \(\rightarrow\) n₁
\(\lambda_{2}=\frac{h c}{E_{2}-E_{1}}\)
\(=\frac{h c}{(-3.4)-(-13.6)}=\frac{h c}{10.20}\)
\(\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\frac{10.20}{1.9}=5.3\)

(i) We know that,
\(\rho=\frac{3 m}{4 \pi R_{0}^{3}}\)
\(\therefore \quad \rho=\frac{3 \times 1.6 \times 10^{-17} \mathrm{~kg}}{4 \times 3.14 \times\left(1.2 \times 10^{-15}\right)^{3}}\)
\(=2.3 \times 10^{17} \mathrm{~kg} / \mathrm{m}^{3}\)
(ii) Nuclear density (P) is independent on mass number, hence nuclear density of a-particle \(\left({ }_{2}^{4} \mathrm{He}\right)\) and thorium \(\left({ }_{90}^{228} \mathrm{Th}\right)\)is equal to each other.
(iii) For a-particle, also nuclear density is equal to 2.3 x 10¹⁷ kg/m ³, as explained earlier.

To produce three lines, the number of excited orbit can be obtained using this relation
\( \frac{n(n-1)}{2} =3 \)
\(n^{2}-3 n =6 \)
\(n^{2}-3 n-6 =0 \)
\((n-3)(n+2) =0 \)
\(n =3 \text { or } n=-2 \)
\(\therefore \quad n =3 . \)
Final energy of electron,
\(E_{n}=\frac{-13.6 \mathrm{eV}}{n^{2}}=\frac{-13.6 \mathrm{eV}}{(3)^{2}}=-1.5 \mathrm{eV}\)

The Rydberg formula is
\(h c / \lambda_{i f}=\frac{m e^{4}}{8 \varepsilon_{o}^{2} h^{2}}\left(\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right)\)
The wavelengths of the first four lines in the Lyman series correspond to transitions from n_i = 2, 3, 4, 5 to n_f = 1. We know that
\(\frac{m e^{4}}{8 \varepsilon_{b}^{7} h^{2}}\) = 13.6 eV = 21.76 × 10^–19 J
Therefore,
\(\lambda _{i 1}=\frac{h c}{21.76 \times 10^{-19}\left(\frac{1}{1}-\frac{1}{n_{i}^{2}}\right)}^{m}\)
\(=\frac{6.625 \times 10^{-34} \times 3 \times 10^{8} \times n_{i}^{2}}{21.76 \times 10^{-19} \times\left(n_{i}^{2}-1\right)} \mathrm{m}=\frac{0.9134 \mathrm{n}_{i}^{2}}{\left(n_{\mathrm{L}}^{2}-1\right)} \times 10^{-7} \mathrm{~m}\)
\(=913.4 n_{i}^{2} /\left(n_{i}^{2}-1\right) \) \(\overset{\unicode{xb0} }{A}\) 
Substituting n_i = 2, 3, 4, 5, we get \(\lambda \)₂₁ = 1218 Å, \(\lambda \)₃₁ = 1028 Å, \(\lambda \)₄₁ = 974.3 Å, and \(\lambda \)₅₁ = 951.4 Å.

Mass of proton, m_p = 1.00783 u
Mass of neutron, m_n = 1.00867 u
In \({ }_7^{14} \mathrm{~N} \text {, }\)there are 7 protons and 7 neutrons.
\(\therefore\) Mass defect, \(\Delta\)m = (7m_p + 7m_n) - m_N
\(=7 \times 1.00783+7 \times 1.00867-14.00307\)
= 0.11243 u
Binding energy of nitrogen nucleus
= \(\Delta\)m \(\times\) 931 MeV
= 0.11243 \(\times\) 931MeV = 104.67 MeV

Atomic mass of \(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\),m₁ = 55.934939 u
\(\begin{array}{l} 26 \\ 56 \end{array} \text { Fe }\) nucleus has 26 protons and (56 − 26) = 30 neutrons
Hence, the mass defect of the nucleus,Δm = 26 x m_H + 30 x m_n − m₁
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm = 26 x 1.007825 + 30 x 1.008665 − 55.934939
= 26.20345 + 30.25995 − 55.934939
= 0.528461 u
But 1 u = 931.5 MeV/c²
∴Δm = 0.528461 x 931.5 MeV/c²
The binding energy of this nucleus is given as:
E_b1 = Δmc²
Where,
c = Speed of light
∴E_b1 = 0.528461 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 492.26 MeV
Average binding energy per nucleon =\(\frac{492.26}{56}=8.79 \mathrm{MeV}\)
Atomic mass of \(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\), m₂ = 208.980388 u
\(\begin{array}{l} 209 \\ 83 \end{array} \text { Bi }\) nucleus has 83 protons and (209 − 83) 126 neutrons.
Hence, the mass defect of this nucleus is given as:
Δm' = 83 x m_H + 126 x m_n − m₂
Where,
Mass of a proton, m_H = 1.007825 u
Mass of a neutron, m_n = 1.008665 u
∴Δm' = 83 x 1.007825 + 126 x 1.008665 − 208.980388
= 83.649475 + 127.091790 − 208.980388
= 1.760877 u
But 1 u = 931.5 MeV/c²
∴Δm' = 1.760877 x 931.5 MeV/c²
Hence, the binding energy of this nucleus is given as:
E_b2 = Δm'c²
= 1.760877 x 931.5 \(\left(\frac{M e V}{c^{2}}\right) \times c^{2}\)
= 1640.26 MeV
Average bindingenergy per nucleon \(=\frac{1640.26}{209}=7.848 \mathrm{MeV}\)

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.

(i) (c)
(ii) (d): \(\omega=\frac{v}{r} . \text { Further } v \propto \frac{1}{n} \text { and } r \propto n^{2},\)
\(\text { Hence } \omega \propto\left(1 / n^{3}\right)\)
(iii) (c)
(iv) (b): The energy of n^th Bohr orbit in hydrogen atom is
\(E_{n}=-\frac{13.6}{n^{2}} \mathrm{eV}\)
For lowest orbit, n = 1
\(\therefore \quad E_{1}=-13.6 \mathrm{eV}\)
Thus, the lowest Bohr orbit in hydrogen atom has the least energy
(v) (b)

(d): The tube light is nothing but a gas discharge tube, which can emit light of different colours. This colour depends mainly upon the nature of the gas inside the tube and the nature of the glass. For neon gas the colour is bright red and for CO₂ it is bluish. Again the fluorescent glow looks yellowish green for soda glass. So it is the nature of the glass and the gas inside the tube which determines the colour of the fluorescent glow. As argon is filled inside a tube light, the colour of the light is white.

(a): Helium-neon laser uses a gaseous mixture of helium and neon. An electric discharge in the gas pumps the helium atoms to higher energy level, (which is meta stable energy level).

Then these helium atom excite the neon atoms to higher level by collision and produce an inverted population of neon atom which emit radiation when they are stimulated to fall to lower level.

(b): When we put R = 10⁷ m^-1and K = 3,4, 5 in the given formula, values of \(\lambda\) calculated lie between 4000 \(\dot A\) and 8000 \(\dot A\), which is the visible region. The reason is true, but does not explain the assertion properly.

(a): Time taken for the light emission from an atom = 10^-8 s.
Time taken for release of energy in fission = 10^-6 s
\(\text { Required ratio }=\frac{10^{-8}}{10^{-6}}=\frac{1}{100}=1: 100\)