CBSE Class 11 Chemistry Structure Of Atom Notes Set 09

These Class 11 CBSE Chemistry notes for Chapter 2 Structure of Atom, Set 09, cover the first half of the chapter: the discovery of the electron, proton and neutron, Thomson's and Rutherford's atomic models, atomic number, mass number and isotopes, the nature of electromagnetic radiation, Planck's quantum theory, the photoelectric effect, atomic spectra and Bohr's model of the hydrogen atom. Six solved examples with full working follow the theory.

Important Notes and Formulas

Discovery of the Sub-atomic Particles

Dalton's atom could not explain why some substances become electrically charged on rubbing. Experiments on electrical discharge through gases at low pressure showed that atoms contain smaller charged particles.

Electron. When a high voltage (about 10,000 V) is applied across two electrodes sealed in a glass tube containing a gas at very low pressure (about 10⁻⁴ atm), invisible rays travel from the cathode to the anode. They are called cathode rays and make a perforated-anode tube glow where they strike a coating of zinc sulphide.

1. Cathode rays travel in straight lines and cast sharp shadows of objects in their path.
2. They are deflected towards the positive plate in an electric field, so they are negatively charged. Stoney named these particles electrons.
3. They turn a light paddle wheel, so they have mass and kinetic energy.
4. Their behaviour does not depend on the material of the electrodes or on the gas in the tube, so electrons are a basic constituent of all matter.
5. They ionise gases and produce X-rays when they strike a metal target.

Charge-to-mass ratio. J. J. Thomson (1897) applied electric and magnetic fields perpendicular to each other and to the path of the electrons. By balancing the two deflections he found e/m = 1.758820 × 10¹¹ C kg⁻¹ for the electron.

Charge on the electron. R. A. Millikan's oil drop experiment showed that the charge on an oil drop is always a whole-number multiple of e, with e = 1.602 × 10⁻¹⁹ C. The mass follows from m = e ÷ (e/m) = 9.109 × 10⁻³¹ kg.

Proton and canal rays. Goldstein used a perforated cathode and found positive rays (canal rays) moving in the opposite direction to cathode rays. They are positive ions formed when fast electrons knock electrons out of gas atoms. Their e/m depends on the gas; it is highest for hydrogen, whose ion is the proton (charge +1.602 × 10⁻¹⁹ C, mass 1.673 × 10⁻²⁷ kg).

Neutron. Chadwick (1932) bombarded beryllium with α-particles and obtained electrically neutral particles of mass 1.675 × 10⁻²⁷ kg, nearly equal to that of the proton.

Atomic Models

Thomson's model (plum pudding model). The atom is a sphere of uniformly spread positive charge, about 10⁻¹⁰ m across, with electrons embedded in it like plums in a pudding or seeds in a watermelon. The atom is electrically neutral and its mass is spread evenly. It fails to explain the results of Rutherford's scattering experiment.

Rutherford's α-particle scattering experiment (1909–1911). A thin gold foil was bombarded with α-particles (He²⁺ nuclei) and the scattered particles were detected on a zinc sulphide screen.

Rutherford's nuclear model. The atom has a tiny, dense, positively charged nucleus holding all the protons and neutrons. Electrons move round it in circular paths, held by electrostatic attraction, like planets round the Sun. The radius of the nucleus is about 10⁻¹⁵ m and that of the atom about 10⁻¹⁰ m, a ratio of about 10⁵.

Drawbacks of Rutherford's model.

1. A revolving electron is accelerated, and by Maxwell's electromagnetic theory it should radiate energy continuously and spiral into the nucleus. Atoms are stable, so the model cannot explain stability.
2. The model does not say how the electrons are arranged around the nucleus or what their energies are, so it cannot explain the line spectra of atoms.

Atomic Number, Mass Number, Isotopes and Isobars

The atomic number Z is the number of protons in the nucleus; in a neutral atom it equals the number of electrons. The mass number A is the total number of protons and neutrons (nucleons). Number of neutrons = A − Z. An atom of element X is written as ^A_ZX.

Isotopes occur in fixed natural proportions, so atomic masses are usually not whole numbers. For example, chlorine is 75% ³⁵Cl and 25% ³⁷Cl, giving an average atomic mass of 35.5 u.

Electromagnetic Radiation

Maxwell showed that light is an electromagnetic wave: oscillating electric and magnetic fields, perpendicular to each other and to the direction of travel. No medium is needed, so the waves travel through a vacuum at c = 3.0 × 10⁸ m s⁻¹.

1. Wavelength (λ): distance between two successive crests or troughs; unit m, nm (10⁻⁹ m) or Å (10⁻¹⁰ m).
2. Frequency (ν): number of waves passing a point per second; unit hertz (Hz = s⁻¹).
3. Wave number (ν̄): number of wavelengths per unit length, ν̄ = 1/λ; unit m⁻¹.
4. Relation: c = νλ.

The electromagnetic spectrum arranges the radiations by wavelength. In order of increasing wavelength (decreasing frequency): γ-rays, X-rays, ultraviolet, visible (about 400–750 nm, violet to red), infrared, microwaves and radio waves. The wave theory could not explain black body radiation or the photoelectric effect.

Planck's Quantum Theory

1. Atoms and molecules emit or absorb energy not continuously but in small discrete packets called quanta (a quantum of light is a photon).
2. The energy of one quantum is proportional to the frequency of the radiation: E = hν = hc/λ, where h = 6.626 × 10⁻³⁴ J s is Planck's constant.
3. Energy is always emitted or absorbed as a whole-number multiple of the quantum: E = nhν, n = 1, 2, 3 ... So energy is quantised.

This explains black body radiation: an ideal body that emits and absorbs all frequencies. As its temperature rises, the emitted radiation shifts to higher frequency (shorter wavelength); a heated iron rod glows first red, then white and then blue-white.

Photoelectric Effect

When light of suitable frequency falls on a metal surface, electrons are ejected. The ejected electrons are called photoelectrons.

1. Electrons are ejected instantly, with no time lag.
2. The number of electrons ejected is proportional to the intensity of the light.
3. Each metal has a threshold frequency ν₀. Below ν₀ no electrons are emitted, however intense the light.
4. Above ν₀, the kinetic energy of the electrons increases with the frequency of the light and does not depend on intensity.

Einstein's explanation. One photon gives all its energy to one electron. Part of it (the work function W₀ = hν₀) is used to escape, and the rest becomes kinetic energy:

hν = hν₀ + ½m_ev², or K.E. = h(ν − ν₀).

Light therefore shows dual behaviour: wave-like (interference, diffraction) and particle-like (black body radiation, photoelectric effect). Energy can also be given in electron volts: 1 eV = 1.602 × 10⁻¹⁹ J.

Atomic Spectra

A continuous spectrum contains all wavelengths of a range (white light through a prism). A line spectrum contains only definite wavelengths, shown as bright lines on a dark background; each element has its own, so it is called the fingerprint of the element.

An emission spectrum is recorded from radiation emitted by excited atoms. An absorption spectrum is obtained by passing white light through a sample: the wavelengths the sample absorbs are missing and appear as dark lines. The study of these spectra is spectroscopy.

Line Spectrum of Hydrogen

The wave number of every line in the hydrogen spectrum is given by the Rydberg formula:

\[ \bar{\nu} = \frac{1}{\lambda} = 1.097 \times 10^{7} \left( \frac{1}{n_1^{2}} - \frac{1}{n_2^{2}} \right) \text{ m}^{-1}, \qquad n_2 > n_1 \]

The number of lines possible when an electron falls from level n to the ground state, by all routes, is n(n − 1)/2. The line of shortest wavelength in a series (n₂ = ∞) is its limiting line.

Bohr's Model of the Hydrogen Atom

1. The electron moves in circular orbits (stationary states) of fixed energy round the nucleus. Energy is neither gained nor lost while it stays in an orbit.
2. Only orbits for which the angular momentum is a whole-number multiple of h/2π are allowed: mvr = nh/2π, n = 1, 2, 3 ...
3. Energy is absorbed or emitted only when the electron jumps between orbits: ΔE = E₂ − E₁ = hν.

For hydrogen and one-electron ions such as He⁺ and Li²⁺ (atomic number Z):

The energy is negative because a free electron at rest at infinity (n = ∞) is taken to have zero energy; an electron bound to the nucleus has lower energy. The most negative energy is at n = 1, the ground state. The ionisation energy of hydrogen is 2.18 × 10⁻¹⁸ J per atom (13.6 eV), and for a one-electron ion it is 13.6 Z² eV.

Limitations of Bohr's model. It fails for atoms with more than one electron, cannot explain the fine structure of the hydrogen spectrum or the splitting of lines in magnetic (Zeeman) and electric (Stark) fields, cannot explain chemical bonding, and ignores the dual nature of matter and the uncertainty principle.

Solved Examples

Example 1. Calculate the number of neutrons in (a) the dipositive zinc ion ⁶⁵Zn²⁺ (Z = 30) and (b) ²⁴Mg²⁺ (Z = 12).
Answer:
(a) Protons = 30; mass number = 65; neutrons = 65 − 30 = 35.
(b) Protons = 12; mass number = 24; neutrons = 24 − 12 = 12.
The charge on an ion changes only the number of electrons, not the neutrons.

Example 2. An ion with mass number 56 carries 3 units of positive charge and has 30.4% more neutrons than electrons. Identify the ion.
Answer:
Let the number of electrons in M³⁺ be x. Neutrons = x + 0.304x = 1.304x.
Protons = electrons in the neutral atom = x + 3.
Mass number: (x + 3) + 1.304x = 56, so 2.304x = 53 and x = 23.
Atomic number = 23 + 3 = 26, so the ion is ⁵⁶₂₆Fe³⁺.

Example 3. The nuclear radius is of the order of 10⁻¹³ cm and the atomic radius 10⁻⁸ cm. Assuming both to be spherical, what fraction of the atomic volume is occupied by the nucleus?
Answer:
Volume of a sphere = (4/3)πr³, so the fraction = (r_nucleus/r_atom)³.
Fraction = (10⁻¹³ ÷ 10⁻⁸)³ = (10⁻⁵)³ = 10⁻¹⁵.
The nucleus occupies only about one part in 10¹⁵ of the atom's volume, which is why most α-particles pass straight through.

Example 4. A light wave has a time period of 2.0 × 10⁻¹⁰ s. Calculate its frequency, wavelength and wave number (c = 3.0 × 10⁸ m s⁻¹).
Answer:
Frequency ν = 1/T = 1 ÷ 2.0 × 10⁻¹⁰ = 5 × 10⁹ s⁻¹.
Wavelength λ = c/ν = 3.0 × 10⁸ ÷ 5 × 10⁹ = 6.0 × 10⁻² m.
Wave number = 1/λ = 1 ÷ 6.0 × 10⁻² = 16.7 m⁻¹.

Example 5. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate the threshold frequency and the work function of the metal.
Answer:
Zero velocity means K.E. = 0, so the photon energy equals the work function and ν = ν₀.
ν₀ = c/λ = 3.0 × 10⁸ ÷ 6800 × 10⁻¹⁰ = 4.41 × 10¹⁴ s⁻¹.
W₀ = hν₀ = 6.626 × 10⁻³⁴ × 4.41 × 10¹⁴ = 2.92 × 10⁻¹⁹ J.

Example 6. Calculate the energy needed to remove the electron from the n = 2 orbit of a hydrogen atom. What is the longest wavelength of light that can do this?
Answer:
Removal means n₁ = 2 and n₂ = ∞: E = 2.18 × 10⁻¹⁸ × (1/2² − 0) = 5.45 × 10⁻¹⁹ J.
For the longest wavelength the photon energy must equal this value exactly: λ = hc/E = 6.626 × 10⁻³⁴ × 3 × 10⁸ ÷ 5.45 × 10⁻¹⁹ = 3.65 × 10⁻⁷ m (365 nm).
Light of shorter wavelength than this (higher energy) will also remove the electron.

Example 7. Which Bohr orbit of Be³⁺ has the same radius as the first orbit of the hydrogen atom?
Answer:
r_n = 52.9 n²/Z pm. For hydrogen, n = 1 and Z = 1: r = 52.9 pm.
For Be³⁺, Z = 4, so 52.9 n²/4 = 52.9, which gives n² = 4 and n = 2.
The second orbit of Be³⁺ has the same radius as the first orbit of hydrogen.

Example 8. Find the speed and the frequency of revolution of the electron in the second Bohr orbit of hydrogen.
Answer:
Speed v₂ = 2.188 × 10⁶ × 1/2 = 1.09 × 10⁶ m s⁻¹.
Radius r₂ = 0.529 × 4 = 2.116 Å = 2.116 × 10⁻¹⁰ m.
Frequency = v/2πr = 1.094 × 10⁶ ÷ (2 × 3.14 × 2.116 × 10⁻¹⁰) = 8.2 × 10¹⁴ s⁻¹.