All matter is made up of tiny particles known as atoms. There are only about 105 different kinds of atoms, and they combine with each other in different ways to form groups called molecules. All matter has been found to be composed of atoms or molecules, and some knowledge of how atoms are made will give us valuable information about the behaviour of matter.
On the basis of his experiments J.J. Thomson proposed a model of internal atomic structure according to which atom consisted of positively charged substance (+ve electric fluid) distributed uniformly over the entire body of the atom, with negative electrons embedded in this continuous positive charge like seeds in a watermelon. It was a good effort to reveal mystery of an atom but it was not the true picture of an atom.
The correct description of the distribution of positive and negative charges within an atom was made in 1911 by a New Zealander when working at Manchester University in England. This was Ernest Rutherford, who was later made Lord Rutherford for his many scientific achievements. He entered into physics during that crucial period of its development when the phenomenon of natural radioactivity had just been discovered, and he was first to realize that radioactivity represents a spontaneous disintegration of heavy unstable atoms.
Rutherford realized that important information about the inner structure of atoms could be obtained by the study of collisions between on rushing α particles and the atoms of various materials forming the target.
The basic idea of the experimental arrangement used by Rutherford in his studies was explained as follows:
a speck of α  emitting radioactive material; a lead shield with a hole that allowed a narrow beam of the α − particles to pass through; a thin metal foil to deflect or scatter them; and a pivoted flourescent screen with a magnifier through which the tiny flashes of light were observed whenever an αparticle struck the screen. The whole apparatus was evacuated, so that the particles would not collide with air molecules.
Most of the αparticles penetrated the foil with very little deflection. An appreciable fraction of them were deflected through large angles  a few were turned back almost as though they had been reflected from the foil. This was a deflection of nearly 180° and a completely impossible phenomenon according to the Thomson's model. Such large deflections required strong forces to be acting, such as those between very smaller charged particles very close together. This would be possible, Rutherford reasoned, if all the positive charge, along with most of the atomic mass, were concentrated in a very small central region which Rutherford called the atomic nucleus. 

Rutherford, knowing the kinetic energy of the αparticles, calculated that they would be within about 1012cm from its centre if α  particles were to be turned back in the direction from which they came.
Because there would be a Coulomb force of attraction between the positive nucleus and the negative electrons, the two would be down together and the atom would vanish unless some provisions were made to prevent it. It was suggested that the electrons might be orbiting rapidly around the nucleus, so that the electrostatic attraction would merely provide the necessary centripetal force.
(i) Rutherford's atomic model was unable to make any predictions about the light that an atom would emit
(ii) More serious than this was its conflict with the accepted laws of electromagnetic theory. An electron revolving rapidly around a nucleus must have a continual centripetal acceleration, and this acceleration would cause a continuous loss of energy by radiation. Bohr calculated that this emission of radiation would cause the electrons in an atom to lose all their energy and fall into the nucleus within a hundred  millionth of a second. Since matter composed of atoms exists permanently, as far as we know, there was obviously something wrong here. Bohr's conclusion was that the conventional classical laws of physics must be wrong, at least when applied to the motion of electrons within an atom.
Bohr in defiance of the well  established laws of classical mechanics and electrodynamics, proposed that the following rules must hold 1. Of all the infinite number of mechanically possible orbits for an electron revolving around a nucleus, only a few are permitted. These are the orbits in which the angular momentum of the electron is an integral multiple of h/2π. 2. While circling around these permitted orbits, the electrons do not emit any electromagnetic radiation, even though conventional electrodynamics holds that they should. 

3. Electrons may jump from one orbit to another, in which case the difference in energy between the two states of motion is radiated as a photon whose frequency is determined by the quantum rule ΔE = hf.
For an electron orbiting in a hydrogen atom, the necessary centripetal force is the electrostatic attraction between the negative electron and the massive, positivelycharged proton, that is the nucleus.
Thus, ke^{2}/r^{2} = mv^{2}/r or r = ke^{2}/mv^{2}(1)
According to Bohr's theory
mvr = nh/2π where, n = 1, 2, 3,….
r = nh/2πmv (2)
From (1) and (2)
v = 2πke^{2} /nh
where h = 6.63 × 1034 J –s (16.1)
Kinetic energyK = 1/2mv2 = (16.2) Potential energyU = −ke2/r= −ke2 (16.3) Total energyE = K + U = − (16.4) Putting the values of k = 9 × 109 Nm2/C2 e = 1.6 × 1019 C and h = 6.63 × 1034 Js, we get E = −1/n^{2} (2.18 × 1018) J = 13.6/n^{2} eV (16.5) 

Radiation and Energy Levels
ΔE = hf
ΔE = E2 − E1
Using equation (16.5)
ΔE = 2.18 × 1018
Applying Planck’s Law,
f = ΔE/h = 3.29 × 1015Hz (16.6)
Dividing the above equation by c = 3 × 108 m/s, we get
m1
or m1 (16.7)
where R∞ = Rydberg constant = 1.097 × 107 m1 .
Bohr showed that Planck's quantum idea were a necessary part of the atom and its inner mechanism; he introduced the idea of quantized energy levels and explained the emission or absorption of radiation as being due to the transition of an electron from one level to another. As a model for even multielectron atoms, the Bohr picture is still useful. It leads to a good, simple, rational ordering of the electrons in larger atoms and qualitatively helps to predict a great deal about chemical behaviour and spectral details.
Bohr's theory is unable to explain the following facts
1. The spectral lines of hydrogen atom is not a single line but a collection of several lines very close together.
2. The structure of multielectron atoms is not explained.
3. No explanation for using the principle of quantisation of angular momentum.
4. No explanation for Zeeman effect
If a substance which gives a line emission spectrum is placed in a magnetic field, the lines of the spectrum get split up into a number of closely spaced lines.
This phenomenon is known as Zeeman effect.
The atom consists of a heavy positively charged nucleus and negatively charged electrons moving around it. The electron is an elementary particle having a mass
me ≈ 9.1 × 10−31 kg and a charge −e, e being an elementary charge approximately equal to
1.60 × 1019C.
The nuclear charge is equal to +Ze, where Z is the atomic number. The atom contains Z electrons, their total charge being −Ze. Consequently, the atom is an electrically neutral system. The size of the nucleus varies depending on Z from 1013 cm to 1012 cm. The size of the atom is a quantity of the order of 108 cm.
The energy of the atom is quantized. This means that it can assume only discrete (i.e. separated by finite gaps) values: E1, E2, E3,…, which are called the energy levels of the atom
(E1 < E2 < E3 < …). Atoms with different Z's have different sets of energy levels.
In a normal (unexcited) state, the atom is on the lowest possible energy level. In such a state, the atom may stay for an infinitely long time. By imparting an energy to the atom, it is possible to transfer it to an excited state with an energy higher than the energy of the ground state. A transition of the atom to a higher energy level may occur as a result of absorption of a photon or as a result of a collision with another atom or a particle, say, an electron.
Excited states of the atom are unstable. The atom can stay in an excited state for about 108 s. After that the atom spontaneously (by its own) goes over to a lower energy level, emitting in this process a photon with an energy
E_{ik} = hf _{ik} (i > k),
where i is the number of the energy level in the initial state and k is the number of the level to which a spontaneous transition of the atom occurred. For example, an atom which is in an excited state with the energy E3 can return to the ground state either directly, by emitting a photon of frequency f31 = (E3 − E1)/h, or through an intermediate state with the energy E2, as a result of which two photons with frequencies f32 = (E3 − E2)/h and f21 = (E2 − E1)/h are emitted.
1. Radius of nth orbit
rn = 0.53 n^{2}/Z Å where Z = atomic number (16.8)
2. Velocity of the electron in the nth orbit
vn = Z/n(c/137) where c = 3 × 108 m/s (16.9)
3. Energy of the electron in the nth orbit
En = 13.6 Z^{2}/n^{2} (eV) (16.10 a)
En = (2.18 × 1018)Z^{2}/n^{2} (J) (16.10 b)
E = K + U (16.11 a )
K = E = U/2 (16.11 b)
U = 2E = 2 K (16.11 c)
4. Wavelength of photon emitted for a transition from n2 to n1
Z^{2}(16.12)
where R∞ = 1.096 × 107 m1 (for a stationary nucleus)
If nucleus is not considered to be stationary
R = R∞/1+m/M(16.13)
where m is the mass of electron and M is the mass of nucleus.
5. Wavelength (Å) of a photon of energy E (eV) is given by
λ = 12400/E(eV)Å (16.14)
6. Momentum of a photon of energy E
p = E/c (16.15)
Example 16.1
A single electron orbits around a stationary nucleus of charge +Ze, where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from second Bohr orbit to the third Bohr orbit.
(a) Find the value of Z
(b) Find the energy required to excite the electron from n = 3 to n = 4
(c) Find the wavelength of radiation required to remove electron from first Bohr’s Orbit to infinity.
(d) Find the kinetic energy, potential energy and angular momentum of the electron in the first Bohr orbit.
Solution
(a) Given ΔE23 = 47.2 eV
Since ΔE = 13.6Z2 eV
∴ 47.2 = 13.6 Z2 (1/2^{2}  1/3^{2}) ⇒ Z = 5
(b) To find ΔE34; n1 = 3; n2 = 4
ΔE = 13.6 Z2 eV
∴ ΔE = 13.6 × 52 (1/3^{2}  1/4^{2}) = 16.53 eV
(c) Ionization energy is the energy required to excite the electron from n = 1 to n = ∞
Thus, ΔE = 13.6 × 52 (1/1^{2} 1/∞^{2}) = 340 eV
The respective wavelength is
λ = hc/ΔE = 12400/ΔE = 12400/340 = 36.47 Å
(d) K = E = +340 eV
U = 2E = 680 eV
L = h/2π= 6.63 x 1034/2π= 1.056 × 1034 Js
Example 16.2
Find the quantum number n corresponding to excited state of He+ ion if on transition to the ground state, the ion emits two photons in succession with wavelengths 108.5 nm and
30.4 nm. The ionization energy of H atom is 13.6 eV.
Solution
The energy transitions for the given wavelengths are
ΔE1 = 12400/λ1 = 12400/1085 = 11.34eV
ΔE2 = 12400/λ2 = 12400/304 = 40.79 eV
Total energy emitted ΔE = ΔE1 + ΔE2 = 52.22 eV
Now ΔE = 13.6 Z2 eV ΔE = energy emitted
or 52.34 = 13.6 × 22 (1/1^{2}  1/n^{2})
Thus, n = 5
Example 16.3
An isolated hydrogen atom emits a photon of 10.2 eV.
(a) Determine the momentum of photon emitted
(b) Calculate the recoil momentum of the atom
(c) Find the kinetic energy of the recoil atom.
[Mass of proton, mP = 1.67 × 1027 kg]
Solution
(a) Momentum of the photon is
p1 = E/c= = 5.44 × 1027 kg m/s
(b) Applying the momentum conservation
p2 = p1 = 5.44 × 1027 kg m/s
(c) K = 1/2 mv2 (v = recoil speed of atom, m = mass of hydrogen atom)
or K =
Substituting the value of the momentum of atom, we get
K = = 8.86 × 1027 J
Example 16.4
A hydrogen atom in a state of binding energy 0.85 eV makes a transition to a state of excitation energy of 10.2 eV. Find the energy and wavelength of photon emitted.
Solution
Since the binding energy is always negative, therefore,
Ei = 0.85 eV
Let ni be the initial binding state of the electron, then
En =
or 0.85 = 13.6
or ni = 4
Binding energy = En = 13.6 Z2/n2
⇒ −0.85 eV = 13.6(1)2/n22 ⇒ n2 = 4
Let the electron now goes to an energy level n whose excitation energy is 10.2 eV. Since the excitation energy ΔE is defined with respect to ground state, therefore
ΔE = 13.6 Z2 eV
or 10.2 = 13.6 × 12
thus nf = 2
So the electron makes a transition from energy level ni = 4 to nf = 2.
Thus, the energy released is ΔE = E4 – E2
or ΔE = 13.6 [1/2^{2}  1/4^{2}]= 2.55 eV
Since λ = hc/ΔE = 12400/2.25eV = 5511 Å
Example 16.5
A particle of charge equal to that of an electron, e and mass 208 times the mass of electron (called a μmeson) moves in a circular orbit around a nucleus of charge +3e (take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system:
(i) Calculate the radius of nth Bohr orbit
(ii) Find the value of n, for which the radius of orbit is approximately the same as that of first Bohr orbit for the hydrogen atom;
(iii) Find the wavelength of radiation emitted when the μmeson jumps from the third orbit to first orbit
(Rydberg’s constant = 1.097 × 107/m)
Solution
(i) Radius of the nth Bohr orbit for hydrogen atom is
rn = 0.53 n^{2}/Z
Since r ∝ 1/m
∴ Radius of nth orbit for μmeson is
rn = 0.53n^{2}/(208)Z
or rn = (8.5 × 104)n2
(ii) (8.5 × 104)n2 = 0.53
∴ n2 = 623
or n ≈ 25
(iii) In case of hydrogen like atom,
ΔE = E3 – E1 = 13.6 Z2 (1  1/3^{2}) = 12.08 eV
since ΔE ∝ m
∴ μmeson, ΔE = (12.8)(208) = 22.6 keV
Thus λ = 12400/ΔE = 12400/22.3 x 10^{3}= 0.548 Å