Electric potential in n^th orbit
\(\mathrm{V} =\mathrm{V}_0 \ln \left(\frac{\mathrm{r}_{\mathrm{n}}}{\mathrm{r}_0}\right)  \)
\(=\mathrm{V}_0\left(\ln \mathrm{r}_{\mathrm{n}}-\ln \mathrm{r}_0\right)  \)
\(=\mathrm{V}_0 \ln \mathrm{r}_{\mathrm{n}}-\mathrm{V}_0 \ln \mathrm{r}_0\)
\(\left|\mathrm{F}_\epsilon\right|=\mathrm{e} \frac{\mathrm{dv}}{\mathrm{dr}}  =\mathrm{e} \frac{\mathrm{d}}{\mathrm{dr}}\left(\mathrm{V}_b / n \mathrm{r}_B-\mathrm{V}_0 / m \mathrm{r}_0\right)  \)
\(=\mathrm{c}\left(\frac{\mathrm{V}_0}{\mathrm{r}_n}-0\right)=\frac{\mathrm{eV}}{\mathrm{r}_{\mathrm{n}}} \)
Centripetal force = coulomb force
\(\frac{m v^2}{r_n}=\mathrm{c} \frac{V_0}{r_n} \Rightarrow v=\sqrt{\frac{e V_0}{m}}=\text { constant }\)
Angular momentum,
\(\mathrm{mvr}_n=\frac{\mathrm{nh}}{2 \pi}\)
\(\mathrm{m}, \mathrm{v}, \mathrm{h}, 2 \pi\) are constants
hence, r_n ∝ n

\(_{ 86 }^{ 220 }{ Rn }\rightarrow _{ 84 }^{ 216 }{ Po }+_{ 2 }^{ 4 }{ He }\)
The mass number of Po = 216
The atomic number of Po = 84
The mass number of He=  4
The atomic number of He = 2

It is defined as the quantity of radiation which produces 1.6 x 10¹² pair of ions in 1 gram of air.

From equation (9.19), the radius of the nuclecus, R = R₀ \(A^\frac13\). Then the volume of the nucleus
\(V=\frac { 4 }{ 3 } \pi { R }^{ 3 }=\frac { 4 }{ 3 } { \pi { R }_{ 0 } }^{ 3 }A\)
By ignoring the mass difference between the proton and neutron, the total mass of the nucleus having mass number A is equal to A.m where m is mass of the proton and is equal to 1.6726 x 10^-27 kg.
Nuclear density.

\(T_{1 / 2}=1600 \text { years }=1600 \times 365 \times 24 \times 60 \times 60 s\)
R = 1 curie = 3.7 x 10¹⁰ Bq, Show that m = 1g
R = λN
Number of atoms Present, N = \(\frac{\mathrm{R}}{\lambda}=\frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2}\)
Mass of 6.023 x 10²³ atoms of  \({ }_{88}^{226} R a=226 g\)
Mass of 1 atom of  \({ }_{88}^{226} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \mathrm{~g}\)
Mass of N atoms of \({ }_{88}^{{ }{266}} \mathrm{Ra}=\frac{226}{6.023 \times 10^{23}} \times \mathrm{Ng}\)
Mass of N atoms of \({ }_{88}^{226} \mathrm{Ra}(\mathrm{m})=\frac{226}{6.023 \times 10^{23}} \times \frac{\mathrm{R}}{0.6931} \mathrm{~T}_{1 / 2} \mathrm{~g}\)
\(\mathrm{m}=\frac{226}{6.023 \times 10^{23}} \times \frac{3.7 \times 10^{10}}{0.6931} \times 1600 \times 365 \times 24 \times 60 \times 60 \mathrm{~g}\)
m = 1.01 g

\(v_{3 \rightarrow 2}=R C\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} R C \)
\(v_{2 \rightarrow 1}=R C\left(\frac{1}{1}-\frac{1}{4}\right)=\frac{3}{4} R C \)
\(v_{3 \rightarrow 1}=R C\left(\frac{1}{1}-\frac{1}{9}\right)=\frac{8}{9} R C \)
\(v_{3 \rightarrow 2}+v_{2 \rightarrow 1}=\frac{5}{36} R C+\frac{3}{4} R C =R C\left(\frac{5}{36}+\frac{3}{4}\right)=R C\left(\frac{5+27}{36}\right) \)
\(v_{3 \rightarrow 2}+v_{2 \rightarrow 1}= R C\left(\frac{32}{36}\right)=\frac{8}{9} R C=v_{3 \rightarrow 1} \)

(i) α - decay :
\(_{ Z }^{ A }{ X\rightarrow }_{ Z-2 }^{ A-4 }{ Y+ }_{ 2 }^{ 4 }{ He }\) 
(ii) β^- decay :
\(_{ Z }^{ A }{ X\rightarrow }_{ Z+1 }^{ A }{ Y+ }{ e }^{ - }+\overset { - }{ v } \)
(iii) β⁺ decay :
\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X} \rightarrow{ }_{\mathrm{Z}-1}^{\mathrm{A}} \mathrm{Y}+\mathrm{e}^{+}+\mathrm{v} \)
(iv) Gamma emission :
\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathrm{X}^{*} \rightarrow{ }_{\mathrm{z}}^{\mathrm{A}} \mathrm{X}+\gamma \text { ray } \)

\(\rho=\frac{\text { mass of the nuclei }}{\text { Volume of the nuclei }}=\frac{\mathrm{Am}}{\frac{4}{3} \pi \mathrm{R}_{0}^{3} \mathrm{~A}}=\frac{\mathrm{m}}{\frac{4}{3} \pi \mathrm{R}_{0}^{3}} \)
\(\rho=\frac{1.67 \times 10^{-27}}{\frac{4}{3} \pi\left(1.2 \times 10^{-15}\right)^{3}}=2.3 \times 10^{17} \mathrm{kgm}^{-3} \)

One atomic mass unit (u) is defined as the (1/12)^th of the mass of the isotope of carbon \(_{ 6 }^{ 12 }{ C }\)  
\(\mathrm{lu}=\frac{\text { mass of }_{6}^{12} \mathrm{C} \text { atom }}{12}=\frac{1.9926 \times 10^{-26}}{12}=1.660 \times 10^{-27} \mathrm{~kg}\)

(i) Minimum energy required to remove an electron from an atom in the ground state is known as binding energy or ionization energy.
(ii) Ionization potential is defined as ionization energy per unit charge.

(i) Most of the alpha particles are un-deflected through the gold foil and went straight.
(ii) Some of the alpha particles are deflected through a small angle.
(iii) A few alpha particles (one in thousand) are deflected through the angle more than 90^o.
(iv) Very few alpha particles returned back (ie) deflected back by 180^o.

Atomic mass unit is defined as \(\frac{1}{12}\) ^th of the mass of one \(_{ 6 }^{ 12 }{ C }\) atom.
E = mc²
= 1.66 x 10-27 x 3 X 108 x 3 x 108 J
= 1.66 x 9 x 10-11 J
= \(\frac { 1.66\times 9\times { 10 }^{ -11 } }{ 1.6\times { 10 }^{ -13 } } \) MeV = 931 MeV.

The difference in masses
Δm = (m_U - m_Th - m_α)
= (232.037156–228.028741 – 4.002603)u
The mass lost in this decay = 0.005812 u
Since 1u = 931MeV, the energy Q released is
Q = (0.005812 u) x (931 MeV / u)
= 5.41 MeV
This disintegration energy Q appears as the kinetic energy of α particle and the daughter nucleus. In any decay, the total linear momentum must be conserved.
Total linear momentum of the parent nucleus = total linear momentum of the daughter nucleus and α particle. Since before decay, the uranium nucleus is at rest, its momentum is zero. By applying conservation of momentum, we get
0 = \({ m }_{ Th }{ \overrightarrow { \upsilon } }_{ Th }+{ m }_{ \alpha }\overrightarrow { \upsilon } _{ \alpha }\)
\({ m }_{ \alpha }\overrightarrow { \upsilon } _{ \alpha }\) = - \({ m }_{ Th }{ \overrightarrow { \upsilon } }_{ Th }\)
It implies that the alpha particle and daughter nucleus move in opposite directions.
In magnitude m_α ሀ_α = m_Th ሀ_Th
The velocity of α particle ሀ_{α =} \(\frac { { m }_{ Th } }{ { m }_{ \alpha } } { \upsilon }_{ Th }\)
Since m_Th > m_α , ሀ_α > ሀ_Th. The ratio of the kinetic energy of α particle to that the daughter nucleus,
\(\frac { K.{ E }_{ \alpha } }{ K.{ E }_{ Th } } =\frac { 1/2{ m }_{ \alpha }{ { \upsilon }_{ \alpha } }^{ 2 } }{ 1/2{ m }_{ Th }{ { \upsilon }_{ Th } }^{ 2 } } \)
By substituting, the value of ሀ_α into the above equation, we get \(\frac { K.{ E }_{ \alpha } }{ K.{ E }_{ Th } } =\frac { { m }_{ Th } }{ { m }_{ \alpha } } =\frac { 228.02871 }{ 4.002603 } =57\)
The kinetic energy of α particle is 57 times greater than the kinetic energy of the daughter nucleus (\(_{ 90 }^{ 228 }{ Th }\))
The disintegration energy Q = total kinetic energy of products
K.E_α + K.E_Th = 5.41 MeV
57K.E_Th + K.E _Th = 5.41 MeV
K.E_Th = \(\frac{5.41}{58}\) MeV = 0.0093 MeV
K.E_α = 57K.E_Th = 57 x 0.093 = 5.301 MeV
In fact, 98% of total kinetic energy is taken by the α particle.

(a) Given that E_n = \(\frac { 54.4 }{ { n }^{ 2 } } eV\)
For n = 1, the ground state energy E₁ = –54.4 eV and for n = 2, E₂ = –13.6 eV. Similarly, E₃ = –6.04 eV, E₄ = –3.4 eV and so on.
For large value of principal quantum number – that is, n = ∞, we get E_∞ = 0 eV.

(b) For a hydrogen-like atom, ground state energy is
E₁ =\(\frac { 13.6 }{ { n }^{ 2 } } { Z }^{ 2 }eV\)
where Z is the atomic number. Hence, comparing this energy with given energy, we get, – 13.6 Z² = – 54.4 ⇒ Z = ±2. Since, atomic number cannot be negative number, Z = 2.
(c) The first excitation energy is
E₁ = E₂ - E₁ = -13.6 eV - (-54.4eV)
= 40.8 eV
Hence, the first excitation potential is
\({ V }_{ 1 }=\frac { 1 }{ e } { E }_{ 1 }=\frac { (40.8eV) }{ e } \)
= 40.8 volt
The first ionization energy is
E_ionization = E_∞ -  E₁ = 0 -(-54.4eV)
= 54.4 eV
Hence, the first ionization potential is
\({ V }_{ ionization }=\frac { 1 }{ e } { E }_{ ionization }=\frac { (54.4eV) }{ e } \)
= 54.4 volt
(d) Consider two photons to be A and B.
Given that photon A with energy 42 eV and photon B with energy 51 eV
From Bohr assumption, difference in energy levels is equal to photon energy, then atom will absorb energy, otherwise, not.
E₂ - E₁ = -13.6eV - (-54.4eV)
= 40.8eV ≈ 41 eV
Similarly,
E₃ - E₁ = -6.04 eV - (-54.4eV)
= 48.36 eV
E₄ - E₁ = -3.4eV - (-54.4eV)
= 51 eV
E₃ - E₂ = -6.04eV - (-13.6eV)
= 7.56 eV
and so on.
But note that E₂ – E₁ ≠ 42 eV, E₃ – E₁ ≠ 42 eV, E₄ – E₁ ≠ 42 eV and E₃ – E₂ ≠ 4₂ eV
For all possibilities, no difference in energy is an integer multiple of photon energy. Hence, photon A is not absorbed by this atom. But for Photon B, E₄ – E₁ = 51 eV, which means, Photon B can be absorbed by this atom
(d) The radius of Bohr orbit is \(r_n=\frac { a_o\times n^2 }{ z }\)
For n = 1, z = 2
\(r_1=\frac { a_o }{ 2 }\)
\(=\frac { 0.529 }{ 2 }\)
= 0.265 Å
(e) Since total energy is equal to negative of kinetic energy in Bohr atom model, we get
\(K{ E }_{ n }=-{ E }_{ n }=-\left( -\frac { 54.4 }{ { n }^{ 2 } } eV \right) \)
\(=-\frac { 54.4 }{ { n }^{ 2 } } eV\)
Since, potential energy is negative of twice the kinetic energy, 
\( U_{ n }=-2K{ E }_{ n }=-2\left( -\frac { 54.4 }{ { n }^{ 2 } } eV \right) \)
\(​​=-\frac { 108.8 }{ { n }^{ 2 } } eV\)
For a ground state, put n = 1
Kinetic energy is KE₁ = 54.4 eV and Potential energy is U₁ = –108.8 eV

Density \(\rho=2.3 \times 10^{17} \mathrm{kgm}^{-3}\), Mass M = 5.97 x 10²⁴ kg
\(\rho=\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}\)
\(R=\left[\frac{M}{\frac{4}{3} \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \mathrm{M}}{4 \pi \rho}\right]^{\frac{1}{3}}=\left[\frac{3 \times 5.97 \times 10^{24}}{4 \times 3.14 \times 2.3 \times 10^{17}}\right]^{\frac{1}{3}}=\left[0.62 \times 10^{7}\right]^{\frac{1}{3}}\)
R = 183.7 m
R ≈180 m

The neutrino has the following properties:
(i) It has zero charge
(ii) It has an antiparticle called anti-neutrino
(iii) Recent experiments showed that the nutrino has very small mass.
(iv) It interacts very weakly with the matter therefore, it is very difficult to detect it.
(v) In every second trillion of neutrino coming from the sun are passing through our body without any interaction.
Initially, it was thought that during beta decay, a neutron in the parent nucleus is converted to the daughter nuclei by emitting only electron as given by
\({ }_{\mathrm{Z}}^{\mathrm{A}} \mathbf{X} \rightarrow \underset{\mathrm{Z}+1}{\mathrm{~A}} \mathbf{Y}+\mathrm{e}^{-}\)
(i) But the kinetic energy of electron coming out of the nucleus did not match with the experimental results.
(ii) In beta decay, the beta particle (i.e, electron) have a continuous range of energies. But the conservation of energy and momentum gives specific single values for electron energy and the recoiling nucleus Y It seems that the conservation of energy, momentum is violated and could not be explained. So beta decay remained as a puzzle for several years.
(iii) But later w. Pauli proposed a third particle which must be present in beta decay to carry away missing energy and momentum.
(iv) Fermi later named this particle the neutrino.
(v) Finally, the neutrino was detected experimentally in 1956 by Fredrick Reines and Clyde Cowan.

(i) When unstable nuclei decay by emitting an \(\alpha\) - particle (\(_{ 4 }^{ 2 }{ He }\) nucleus), it loses two protons and two neutrons. As a result, its atomic number Z decreases by 2, the mass number decreases by 4. 
\(_{ Z }^{ A }{ X }\rightarrow _{ Z-2 }^{ A-4 }{ Y+ }_{ 2 }^{ 4 }{ He }\)
(ii) X is called the parent nucleus and Y is called the daughter nucleus.
Example : when uranium \(_{ 92 }^{ 238 }U\) emit (α - particle) it is converted into thorium  \(_{ 90 }^{ 234 }{ Th }\) .
\(_{ 92 }^{ 238 }{ U\rightarrow }_{ 90 }^{ 234 }{ Th }+_{ 2 }^{ 4 }{ He }\)
(iii) Total mass of the daughter nucleus and \({ }_{2}^{4} \mathrm{He}\) nucleus is always less than that of parent nucleus.
(iv) The difference in mass \(\left(\Delta m=m_{X}-m_{Y}-m_{\alpha}\right)\) is released as energy called disintegration energy Q 
\(\mathrm{Q}=\Delta \mathrm{m} \times \mathrm{c}^{2}=\left(\mathrm{m}_{\mathrm{X}}-\mathrm{m}_{\mathrm{Y}}-\mathrm{m}_{\alpha}\right) \mathrm{c}^{2}\) 
(v) For spontaneous decay (natural radioactivity) Q > 0. In alpha decay process.
(a) The disintegration energy is positive (Q > 0).
(i) if the parent nucleus is not at rest, Q is equal to the net kinetic energy gained in the decay process.
(ii) if the parent nucleus is at rest, Q is equal to the total kinetic energy of daughter nucleus and the \({ }_{2}^{4} \mathrm{He}\) nucleus.
(b) The disintegration energy is negative (Q < 0).
The decay process cannot occur spontaneously and energy must be supplied to induce the decay.

The average binding energy per nucleon is the energy required to separate single nucleon from the particular nucleus.


\(\overline{\mathrm{BE}}=\frac{\left[\mathrm{Zm}_{\mathrm{H}}+\mathrm{Nm} _{\mathrm{n}}-\mathrm{M}_{\mathrm{A}}\right] \mathrm{c}^{2}}{\mathrm{~A}}\)
\(\overline { BE } \) is plotted against A of all known nuclei.
Important inferences from the average binding energy curve:
(i) The value of \(\overline { BE } \) rises as the mass number increases until it reaches a maximum value of 8.8 MeV for A = 56 (iron) and then it slowly decreases.
(ii) The average binding energy per nucleon is about 8.5 MeV for nuclei having mass number between A = 40 and 120. These elements are comparatively more stable and not radioactive.
(iii) For higher mass numbers, the curve reduces slowly and \(\overline { BE } \) for uranium is about 7.6 MeV. They are unstable and radioactive.
(iv) From Figure, If two light nuclei with A<28 combine with a nucleus with A<56, the binding energy per nucleon is more for final nucleus than initial nuclei. Thus, if the lighter elements combine to produce a nucleus of medium value A, a large amount of energy will be released. This is the basis of nuclear fusion and is the principle of the hydrogen bomb.
(v) If a nucleus of heavy element is split (fission) into two or more nuclei of medium value A, the energy released would again be large. The atom bomb is based on this principle and huge energy of atom bombs comes from this fission when it is uncontrolled.

The electrostatic force is a conservative force, the potential energy for the electron in n^th orbit is
\(U_{n} =\frac{1}{4 \pi \varepsilon_{0}} \frac{(+Z e)(-e)}{r_{n}}=-\frac{1}{4 \pi \varepsilon_{0}} \frac{Z^{2}}{r_{n}} \)      \(\left[ \because r_n=\frac{\varepsilon_{0} h^{2} n^{2}}{\pi m Z e^{2}}\right]\)
\(U_{n} =-\frac{1}{4\varepsilon_{0}} -\frac{Z^{2} \mathrm{me}^{4}}{h^{2} n^{2}} \)
The kinetic energy of electron in n^th orbit is
\(\mathrm{KE}_{\mathrm{n}}=\frac{1}{2} \mathrm{mv}_{\mathrm{n}}^{2}=\frac{\mathrm{Z}^{2} m \mathrm{e}^{4}}{8 \varepsilon_{0}^{2} \mathrm{~h}^{2} \mathrm{n}^{2}}\)
This implies that U_n = -2KE_n
Total energy of electron in the n^th orbit is
\(E_{n}=K E_{n}+U_{n}=K E_{n}-2 K E_{n}=-K E_{n} \)
\(E_{n}=-\frac{Z^{2} m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}} \)
For Hydrogen atom Z = 1
\(E_{n}=-\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2} n^{2}} \text { joule }\)
n - principal quantum number
The negative sign indicates that the electron is bound to the nucleus.
Substituting the values of mass and charge of an electron (m and e), permittivity of free space \(\varepsilon^{0}\)  and Planck's constant h and expressing in terms of (+(eV)), we get
\(E_{n}=-13.6\left(\frac{1}{n^{2}}\right) e V\)
(i) For the first orbit (ground state), the total energy of electron is E₁ = - 13.6 eV.
(ii) For the second orbit (first excited state), the total energy of electron is E₂ = -3.4 eV.
(iii) For the third orbit (second excited state), the total energy of electron is E₃ = -1.51 eV and so on.

According to Einstein mass - energy equivalence, E = mc²
Here \(m=l u=1.66 \times 10^{-27} \mathrm{~kg} \)
\(E =1 \mathrm{u} \times \mathrm{c}^{2}=1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J} \)
\(E =\frac{1.66 \times 10^{-27} \times\left(3 \times 10^{8}\right)^{2}}{1.6 \times 10^{-19}} \mathrm{eV}=931 \times 10^{6} \mathrm{eV}=931 \mathrm{MeV} \)

(i) The electron in an atom moves around nucleus in circular orbits under the influence of Coulomb electrostatic force of attraction. This Coulomb force gives necessary centripetal force for the electron to undergo circular motion.
(ii) Electrons in an atom revolve around the nucleus only in certain discrete orbits called stationary orbits where it does not radiate electromagnetic energy. Only those discrete orbits allowed are stable orbits.
(iii) The angular momentum of the electron in these stationary orbits are quantized (ie) L = \(\frac{nh}{2\pi}\) This is known as Bohr quantization condition.
(iv) The energy of the orbits are not continuous but only discrete. This is called quantization of energy.
(v) An electron can jump from one orbit to another orbit by absorbing or emitting a photon whose energy is equal to the difference in energy between the two orbital levels.

Total rest mass (initial) = 236.0526 u
Total rest mass (final) = 235.8293 u
Decrease in rest mass due to fission = 0.2233 u
Energy released
= 0.2233 u x 931 \(\frac{MeV}{u}\) = 207.9 MeV.

a) \(_{ 1 }^{ 2 }{ D+ }_{ 1 }^{ 2 }{ D }\rightarrow _{ 1 }^{ 3 }{ T+ }_{ 1 }^{ 1 }{ H }+Q\)
∴ Q₁ = 0.00433 x 931 MeV
= 4.031 MeV.
\(_{ 1 }^{ 3 }{ T }+_{ 1 }^{ 2 }{ D }\rightarrow _{ 2 }^{ 4 }{ He+ }_{ 0 }^{ 1 }{ n }+{ Q }_{ 2 }\)
∴ Q₂ = 0.01888 x 931 MeV
= 17.577 MeV.
(b) Energy released per deuterium nucleus
\(=\frac { 2161 }{ 3 } =7.203MeV\)
(c) Percentage of the rest mass of deuterium released
\(=\frac { 7.203 }{ (2.014102\times 931)MeV } =0.384 \%\) 

(i) The smoke detector uses around 0.2 mg of a man-made weak radioactive isotope called americium \((_{ 95 }^{ 241 }{ Am })\)

(ii) This radioactive source is placed between two oppositely charged metal plates and α radiations from \(_{ 95 }^{ 241 }{ Am }\) continuously ionize the nitrogen, oxygen molecules in the air space between the plates
(iii) As a result, there will be a continuous flow of small steady currents in the circuit.
(iv) If smoke enters, the radiation is being absorbed by the smoke particles rather than air molecules.
(v) As a result, the ionization and along with it the current is reduced. This drop-in current is detected by the circuit and the alarm starts.
(vi) The radiation dosage emitted by americium is very much less than the safe level, so it can be considered harmless.

Rutherford atom model helps in the calculation of the diameter of the nucleus and also the size of the atom but has the following limitations
(a) This model fails to explain the distribution of electrons around the nucleus and also the stability of the atom.
According to classical electrodynamics,.any accelerated charge emits electromagnetic radiations. Due to emission of radiations, it loses its energy. Hence, it can no longer .sustain the circular motion. The radius of the orbit, therefore, becomes smaller and smaller (undergoes spiral motion) and finally the electron should fall into the nucleus and the atoms should disintegrate. But this does not happen.

(b) According to this model, emission of radiation must be continuous and must give continuous emission spectrum but experimentally we observe only line (discrete) emission spectrum for atoms.

(i) The process of breaking up of the nucleus of a heavier atom into two smaller nuclei with the release of a large amount of energy is called nuclear fission.
Examples :
\(_{ 92 }^{ 235 }{ U+ }_{ 0 }^{ 1 }{ n }\rightarrow _{ 92 }^{ 236 }{ { U }^{ * } }\rightarrow _{ 56 }^{ 141 }{ Ba }+_{ 36 }^{ 92 }{ Kr }+3_{ 0 }^{ 1 }{ n }+Q\)
\(_{ 92 }^{ 235 }{ U+ }_{ 0 }^{ 1 }{ n }\rightarrow _{ 92 }^{ 236 }{ { U }^{ * } }\rightarrow _{ 54 }^{ 140 }Xe+_{ 38 }^{ 94 }{ Sr }+2_{ 0 }^{ 1 }{ n }+Q\)
(ii) When the slow neutron is absorbed by the uranium nuclei, the mass number increases by one and goes to an excited state \(_{ 92 }^{ 235 }{ U }\).
(iii) But this excited state does not last longer than 10^-12s and decay into two daughter nuclei along with 2 or 3 neutrons.
Energy released in fission :
(i) We can calculate the energy (Q) released in each uranium fission reaction. We choose the most observed fission reaction which is given in the equation.

\(_{ 92 }^{ 235 }{ U+ }_{ 0 }^{ 1 }{ n }\rightarrow _{ 92 }^{ 236 }{ { U }^{ * } }\rightarrow _{ 56 }^{ 141 }{ Ba }+_{ 36 }^{ 92 }{ Kr }+3_{ 0 }^{ 1 }{ n }+Q\)
Mass of \({ }_{92}^{235} \mathrm{U}\) = 235.045733 u
Mass of \(_{ 0 }^{ 1 }{ n }\) = 1.008665 u
Total mass of reactants = 236.054398 u
Mass of \(_{ 56 }^{ 141 }{ Ba }\) = 140.9177 u
Mass of \(_{ 92 }^{ 36 }{ Kr }\) = 91.8854u
Mass of 3 neutrons = 3.025995 u
The total mass of products = 235.829095 u
Mass detect m =236.054398 u - 235.829095 u
= 0.225303u
So the energy released in each fission
= 0.225303 x 931 MeV = 200.MeV
(ii) This energy first appears as kinetic energy of daughter nuclei and neutrons. But later, this kinetic appears in the form of heat given to the surrounding.
Properties :
(i) The fission is accompanied by the release of neutrons. From each reaction, on an average, 2.5 neutrons are emitted.
(ii) The energy released in the nuclear fission is many times greater than the energy released in chemical reactions.
(iii) Energy released per fission is 200 MeV.

The motion of oil drop inside the chamber can be controlled by adjusting electric field. The oil drop can be moved up or down or even kept balanced in the field of view for sufficiently long time.
Construction:
(i) The apparatus consists of two horizontal circular metal plates A and B each with diameter around 20 cm and are separated by a small distance 1.5 cm.
(ii) These two parallel plates are enclosed in a chamber with glass walls.
(iii) Plates A and B are given a high potential difference around 10 kV such that electric field acts vertically downward
(iv) A small hole is made at the center of the upper plate A.
(v) Atomizer is kept above the hole to spray the liquid.
Working:
(i) When a fine droplet of highly viscous liquid (like glycerine) is sprayed using atomizer, it falls freely downward through the hole under the influence of gravity alone.
(ii) Few oil drops in the chamber can acquire electric charge (negative charge) because of friction with air or passage of x-rays in between the parallel plates.
(iii) The chamber is illuminated by light and oil drops can be seen clearly using microscope.
(iv) These drops can move either upwards or downward.
(v) Let m be the mass of the oil drop and q be its charge. Then the forces acting on the droplet are
(a) gravitational force F_g = mg
(b) electric force F_e = qE
(c) buoyant force F_b
(d) viscous force F_v
(a) Determination of radius of the droplet:
(i) When the electric field is switched off, the oil drop accelerates downwards. Due to presence of air drag forces, the oil drops attain its terminal velocity and moves with constant velocity.
(ii) This velocity can be measured by finding the time taken by the oil drop to fall through a predetermined distance.

(iii) From the free body diagram, we note that viscous force and buoyant force (upward) balance the gravitational force (downward).
(iv) Let us assume that oil drop to be spherical in shape.
(v) Let p be the density of the oil drop, and r be the radius of the oil drop, then the mass of the oil drop, the oil drop can be expressed in terms of its density as, \( \rho=m/v \Rightarrow m=\left(\frac{4}{3} \pi r^{3}\right) \rho\)
∵ (Volume of the sphere, V = \(\frac{4}{3} \pi r^{3})\)
Then gravitational force \(\mathrm{F}_{\mathrm{g}}=\mathrm{mg}=\left(\frac{4}{3} \pi \mathrm{r}^{3}\right) \rho g\)
(vi) Let σ be the density of air, the upthrust force experienced by the oil drop due to displaced air is \(F_{b}=\left(\frac{4}{3} \pi r^{3}\right) σ g\)
(vii) Once the oil drop attains a terminal velocity v, the net downward force acting on the oil drop is equal to the viscous force acting opposite to the direction of motion of the oil drop. From Stokes law, the viscous force on the oil drop is
\(\mathrm{F}_{\mathrm{v}}=6 \pi \eta \mathrm{rv}\)
(ix) From free body diagram, the force balancing equation is,
\(F_{g}=F_{b}+F_{v} \)
\(\left(\frac{4}{3} \pi r^{3}\right) \rho g=\left(\frac{4}{3} \pi r^{3}\right) \sigma g+6 \pi \eta r v \)
\(\frac{4}{3} \pi r^{3}(\rho-\sigma) g=6 \pi \eta r v \)
\(\frac{2}{3} r^{2}(\rho-\sigma) g=3 \eta v \)
Hence radius of the oil drop is \(r=\left[\frac{9 \eta v}{2(\rho-\sigma) g}\right]^{\frac{1}{2}} \ldots\)               ........(1)
(b) Determination of electric charge:
(i) Now switch on the electric field.
(ii) Upward electric force on charged oil drops is qE.
(iii) Choose any one drop in the field of view of microscope.
(iv) Strength of the electric field is adjusted to make that particular drop to be stationary.
(v) Being oil drop is at rest, the viscous force acting on the oil drop is zero.
(vi) Then, from the free body diagram.
\(F_{g}=F_{b}+F_{v} \)
\(\left(\frac{4}{3} \pi r^{3}\right) \rho g=\left(\frac{4}{3} \pi r^{3}\right) \sigma g+q E \)
\(\frac{4}{3} \pi r^{3}(\rho-\sigma) g=q E \)
\(q=\frac{4}{3 E} \pi r^{3}(\rho-\sigma) g \ldots \ldots \ldots \ldots \ldots \) (2)
Substituting (1) in (2)
\(q=\frac{18 \pi}{E}\left(\frac{\eta^{3} v^{3}}{2(\rho-\sigma) g}\right)^{\frac{1}{2}}\)
(vii) Millikan repeated this experiment several times and computed the charges on oil drops. He found that the charge of any oil drop can be written as integral multiple of a basic value, -1.6 x 10^-19 C which is nothing but the charge of an electron.