FUNDAMENTAL PARTICLES

Atoms are made up-essentially, of three fundamental particles, which differ in mass and electric charge as follows:

The atomic mass unit (amu) is 1/12th of the mass of an individual atom of ₆C¹², i.e.

1.660565 x 10^-27 kg. The neutron and proton have approximately equal masses of 1 amu and the electron is about 1836 times lighter; its mass can sometimes be neglected as an approximation. The electron and proton have equal, but opposite, electric charges; the neutron is not charged.

The existence of electrons in atoms was first suggested, by J.J. Thomson, as a result of experimental work on the conduction of electricity through gases at low pressures, which produces cathode rays and x-rays, and a study of radioactivity by Becquerel, the Curies and Rutherford.

An atom is electrically neutral, and if it contains negatively charged electrons it must also contain some positively charged particles, and the supposition that they existed within atoms came about as a result of Rutherford's experiments in which he bombarded elements with the $\alpha$. - rays and

$\alpha$-rays were given off by radioactive elements. The neutron was discovered in 1932 by James Chadwick by bombarding beryllium with $\alpha$ rays.

DISCOVERY OF ELECTRON: CATHODE RAYS

During the latter half of the nineteenth century, it was found that while normally dry gases do not conduct an electric current, they do so under very low pressure and then patches of light are seen. The passage of electricity through gases as studied by a number of physicists, particularly by Faraday, Davy, Crookes and J.J. Thomson.

When a current of high voltage (10,000 volts) is passed through a gas of air kept at a very low pressure (0.01 – 0.03 mm) blue rays are seen emerging from the case. These rays are called "Cathode Rays".

Some of the important properties of the cathode rays studied by Sir J.J. Thomson and others are given below:

$\bullet$Cathode rays come out at right angles to the surface of the cathode and move in

straight lines.

$\bullet$Their path is independent on the position of the anode.

$\bullet$They produce phosphorescence on certain salts like ZnS and fluorescence on glass.

$\bullet$They blacken photographic plates.

$\bullet$The rays pass through thin sheet of metals. If the metal sheet is too thick to be penetrated the rays cast a shadow.

$\bullet$They produce X-ray when they strike a metal.

$\bullet$The rays ionize a gas through which they pass.

$\bullet$They heat a substance on which they fall.

$\bullet$They rotate a light wheel placed in their paths. This shows that cathode rays contain material particles having both mass and velocity.

$\bullet$The mass of a particle present in cathode rays is found to be 1/1837 of H-atom. This shows that the particle is of sub-atomic nature.

$\bullet$Cathode rays are deflected by a magnetic or an electric field showing the particle to be electrically charged, the direction of deflection shows that they are negatively charged.

$\bullet$Cathode rays contain the smallest unit of negative charge.

$\bullet$No cathode ray was produced when the tube was completely evacuated.

$\bullet$Different gases produce same cathode rays as they have the same e/m (charge/mass) ratio. This indicates that the particles present in cathode rays are fundamental constituent of all matter.

Sir J.J. Thomson named these negatively charged sub-atomic particles as electron.

"A sub-atomic particle which is a fundamental constituent of all matter having a mass 1/1837^th of a H-atom and which carries the smallest unit of negative charge is called an electron".

Determination of Velocity and Charge/mass (e/m) ratio of Electrons:

Sir J.J. Thomson (1897) extended the cathode ray experiment for the determination of velocity of electrons and their charge/mass ratio, The value of e/m for an electron = 1.76 x 10⁸ C/g.

For the H⁺ ion (proton), e/m = 96500/1.008 C/g.

Millikan's Oil Drop Method: Determination of Charge on an Electron:

In 1909, Millikan measured the charge on an electron by his oil drop method. In this method a spray of oil droplets is produced by an atomizer, some of which pass through an opening into a viewing chamber, where we can observe them with a microscope. Often these droplets have an electric charge, which is picked up from the friction forming the oil droplets. A droplet may have one or more additional electrons in it, giving it a negative charge.

As the droplet falls to the bottom of the chamber, it passes between two electrically charged plates. The droplet can be suspended between them; we adjust the voltage in the plates so that the electrical attraction upward just balances the force of gravity downward. We then use the voltage needed to establish this balance to calculate the mass - to charge ratio for the droplet. Because we already know the mass of the droplet we can find the charge on it.

Millikan's found that the charge on all droplets could be expressed as whole number multiples of e, where the value of e is 1.602x10^-19 C. By combining e/m. ratio and 'e' we calculate mass of the electron

$M_e=\frac{e}{e/m}=\frac{1.6022\times {10}^{-19}}{1.76\times {10}^8}$

= 9.104x10^-31 kg

This very small value shows that the electron is a subatomic particle. Thus charge on an electron = 1.602 x 10^–19C.

DISCOVERY OF PROTON: POSITIVE RAYS OR CANAL RAYS

Atoms are electrically neutral. Hence after the discovery of the negatively charged constituent (electron) of an atom, attempts were made to discover the positively charged counterpart of electrons. By using a discharge tube containing a perforated cathode. Goldstein (1886) found that some rays passed through these holes in a direction opposite to that of the cathode rays.

These are called the positive rays or canal rays. J.J. Thomson (1910) measured their charge by mass ratio from which he was able to deduce that these contain positive ions. Their

properties are:

$\bullet$They are positively charged.

$\bullet$The positive charge is either equal to or whole number multiple of the charge on an electron.

$\bullet$When hydrogen gas was filled in the discharge tube the positive charge on the positive rays was equal to the negative charge on an electron, and the mass was less than the hydrogen atom.

$\bullet$Unlike cathode rays the properties of positive rays are characteristics of the gas in

the tube.

$\bullet$The deflection of positive rays under the influence of an electric or magnetic field is smaller than that of the cathode rays for the same strength of field. This shows that the positive rays have a greater mass than that of electrons.

$\bullet$The mass of the positive rays depends on the atomic weights or molecular weights of the gases in the discharge tube. The charge/mass ratio also varies because the change in positive charge on the rays. It may be either equal to or integral multiple of the charge on an electron.

$\bullet$The lightest of all particles identified in positive rays from different elements was one with a mass very slightly less than that of hydrogen atom (or nearly equal to H-atom). The lightest positively charged particle is called a proton (P or P⁺). Positive rays are atomic or molecular resides from which some electrons have been removed. The removed electrons constitute the cathode rays and the positive residues form the positive or canal rays.

The mass of a proton is very slightly less than that of a H-atom. This shows that protons are sub-atomic particles. Protons are fundamental constituent of matter because positive rays are produced by all substances.

"A sub-atomic particle, which is a fundamental constituent of all matter having a mass slightly less than that of H-atom and which carries a positive charge equal in magnitude to the charge on an electron, is called a proton". A proton is denoted by p or p⁺ of ₊₁p.

Comparison of Positive (Canal) Rays and Cathode Rays:

DISCOVERY OF NEUTRON

After the discovery of electrons and protons. Rutherford (1920) had predicted the existence of a neutral fundamental particle. In 1932, Chadwick bombarded the element Beryllium with $\alpha$-particles and noticed the emission of a radiation having the following characteristics.

$\bullet$The radiation was highly penetrating.

$\bullet$The radiation was unaffected by magnetic and electric fields which show that it is electrically neutral.

$\bullet$It was found to have approximately the same mass as the protons.

The name 'neutron' was given to this sub-atomic particle. It is denoted by n or${}_o^{1}n$. Bombardment of beryllium by $\alpha$-particles results in the formation of carbon and neutrons are emitted.

Be $\stackrel{\alpha }{\to }C{+}_0^{1}n$

At present there are a number of evidences which confirm that like electron, proton and neutron is also a fundamental constituent of atoms (a single exception is ${}_1^{1}H$atom which does not contain any neutron)

Mass of a neutron is 1.008930 amu (1.6753 x 10^–24g or 1.6753 x 10^–27 kg)

Neutron "A sub-atomic particle, which is a fundamental constituent of matter having mass approximately equal to the hydrogen atom and which is electrically neutral, is called a neutron".

Illustration 1. The neutron is attracted towards

(A) Positive charged particles

(B) Negative charged particles

(C) Not attracted by any charge

(D) None of these

Solution: Neutron is an uncharged particle. Hence (C) is correct.

ATOMIC TERMS

Nuclide: Various species of atoms in general.

Nucleons: Sub-atomic particles in the nucleus of an atom, i.e., protons and neutrons.

Isotopes: Atoms of an element with the same atomic number but different mass number.

Mass number (A): Sum of the number of protons and neutrons, i.e., the total number of nucleons,

Atomic number (Z): The number of protons in the nucleus of an atom. This, when subtracted from A, gives the number of neutrons.

Isobars: Atoms, having the same mass numbers but different atomic numbers, e.g.. ₁₅P³² and ₁₆S³².

Isotones: Atoms having the same number of neutrons but different number of protons or mass number, e.g., ${}_6^{14}C,{}_8^{16}O,{}_7^{15}N.$

Isoelectronic species: Atoms molecules or ions having the same number of electrons, e.g., N₂, CO, CN^-.

Nuclear isomers: Atoms with the same atomic and mass numbers but different radioactive properties, e.g., uranium X (half life 1.4 min) and uranium Z (half life 6.7 hours).

Atomic mass unit: Exactly equal to 1/12th of the mass of ₆C¹² atom.

(a.m.u.): 1 a.m.u. = 1.66 x 10^–24 g » 931.5 MeV

Illustration 2. The ion that is isoelectronic with CO is

(A) CN^- (B) O₂⁺

(C) O₂^- (D) N₂⁺

Solution: Both CO and CN^- have 14 electrons. Hence (A) is correct

ATOMIC MODELS

We know the fundamental particles of the atom. Now let us see, how these particles are arranged in an atom to suggest a model of the atom.

Thomson's Model:

J.J. Thomson, in 1904, proposed that there was an equal and opposite positive charge enveloping the electrons in a matrix. This model is called the plum – pudding model after a type of Victorian dessert in which bits of plums were surrounded by matrix of pudding.

This model could not satisfactorily explain the results of scattering experiment carried out by Rutherford who worked with Thomson.

Rutherford's Model:

$\alpha$– particles emitted by radioactive substance were shown to be dipositive Helium ions (He⁺⁺) having a mass of 4 units and 2 units of positive charge.

Rutherford allowed a narrow beam of $\alpha$–particles to fall on a very thin gold foil of thickness of the order of 0.0004 cm and determined the subsequent path of these particles with the help of a zinc sulphide fluorescent screen. The zinc sulphide screen gives off a visible flash of light when struck by an $\alpha$ particle, as ZnS has the remarkable property of converting kinetic energy of $\alpha$ particle into visible light. [For this experiment, Rutherford specifically used $\alpha$ particles because they are relatively heavy resulting in high momentum].

Observation:

$\bullet$Majority of the $\alpha$–particles pass straight through the gold strip with little or no deflection.

$\bullet$Some $\alpha$–particles are deflected from their path and diverge.

$\bullet$Very few $\alpha$–particles are deflected backwards through angles greater than 90°.

$\bullet$Some were even scattered in the opposite direction at an angle of 180° [Rutherford was very much surprised by it and remarked that "It was as incredible as if you fired a 15­–inch shell at a piece of tissue paper and it came back and hit you"]. There is far less difference between air and bullet than there is between gold atoms and $\alpha$-particle assuming of course that density of a gold atom is evenly distributed. The distance of nucleus from where the $\alpha$ - particle returns back through 180° is called distance of closet approach and is given by

$r_0=\frac{q_1{q}_2}{4\pi {\epsilon }_0\left(\frac{1}{2}m{v}^2\right)}$

Conclusions:

$\bullet$The fact that most of the $\alpha$ - particles passed straight through the metal foil indicates the most part of the atom is empty.

$\bullet$The fact that few $\alpha$ - particles are deflected at large angles indicates the presence of a heavy positively charge body i.e., for such large deflections to occur $\alpha$ - particles must have come closer to or collided with a massive positively charged body.

$\bullet$The fact that one in 20,000 have deflected at 180° backwards indicates that volume occupied by this heavy positively charged body is very small in comparison to total volume of the atom.

Illustration 3. An $\alpha$ - particle is traveling towards gold nuclei returns back through 10^-10 m from it. What is the velocity of the $\alpha$ - particle. [Given 1 amu = 1.66 x 10^-27 kg, atomic mass of He = 4 and gold = 79 and $\frac{1}{4\pi {\epsilon }_0}$ = 9 x 10⁹ Nm²C^-2]

Solution: We known that $r_o=\frac{q_1{q}_2}{4\pi {\epsilon }_0\left(\frac{1}{2}m{v}^2\right)}$

$\therefore v^2=\frac{2q_1{q}_2}{4\pi {\epsilon }_{0}m{r}_0}$

Now one He atom has charge (q₁) = 2e

One gold atom has charge (q₂) = 79e

Putting these values we get

$v^2=\frac{2\times \left(2\times 1.6\times {10}^{-19}\right)\times \left(79\times 1.6\times {10}^{-19}\right)\times 9\times {10}^9}{4\times 1.66\times {10}^{-27}\times {10}^{-10}}$

$\therefore$ v = 3.311 x 10⁵ m/s

Conclusions of $\alpha$-Scattering Experiment:

On the basis of the above observation, and having realized that the rebounding $\alpha$-particles had met something even more massive than themselves inside the gold atom, Rutherford proposed an atomic model as follows.

$\bullet$All the +ve charge and nearly the total mass of an atom is present in a very small region at the centre of the atom. The atom's central core is called nucleus.

$\bullet$The size of the nucleus is very small in comparison to the size of the atom. Diameter of the nucleus is about 10^–13cm while the atom has a diameter of the order of 10^–8 cm. So, the size of atom is 10⁵ times more than that of nucleus.

$\bullet$Most of the space outside the nucleus is empty.

$\bullet$The electrons, equal in number to the net nuclear positive charge, revolve around the nucleus with fast speed just like planets around the sun.

$\bullet$The centrifugal force arising due to the fast speed of an electron balances the coulombic force of attraction of the nucleus and the electron remains stable in its path. Thus according to him atom consists of two parts (a) nucleus and (b) extra nuclear part.

Defects in Rutherford's Atomic Model:

$\bullet$Position of electrons: The exact positions of the electrons from the nucleus are not mentioned.

$\bullet$Stability of the atom: Bohr pointed out that Rutherford's atom should be highly unstable. According to the law of electro–dynamics, when a charged body moves under the influence of an attractive force, it loses energy continuously in the form of electromagnetic radiation. The electron should therefore, continuously emit radiation and lose energy. As a result of this a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus.

It was calculated that the electron should fall into the nucleus in less than 10^–8 sec. But it is known that electrons keep moving outside the nucleus.

To solve this problem Neils Bohr proposed an improved form of Rutherford's atomic model.

Before going into the details of Neils Bohr model we would like to introduce you some important atomic terms.

Illustration 4. Prove that density of the nucleus is constant.

Solution: Radius of the nucleus = 1.33 x 10^–13 x A^1/3 cm, where A is the mass number

= 1.33 x 10^–11 x A^1/3 m

Density of nucleus = $\frac{Mass}{Volume}$

=$\frac{A\times 1.66\times {10}^{-27}\overrightarrow{\leftarrow }kg}{\frac{4}{3}\times \pi \times {(1.33\times {10}^{-11}\times A^{1/3})}^3}$= $\frac{1.66\times {10}^{-27}\times 3}{4\times 3.14\times {(1.33\times {10}^{-11})}^3}$kg/m³ = constant

Thus density of nucleus is constant, independent of the element under consideration.

SOME IMPORTANT CHARACTERISTICS OF A WAVE

A wave is a sort of disturbance which originates from some vibrating source and travels outward as a continuous sequence of alternating crests and troughs. Every wave has five important characteristics, namely, wavelength ($\lambda$), frequency ($\nu$), velocity (c), wave number $\left(\bar{\nu }\right)$ and amplitude (a).

Electronic Magnetic Radiation:

Ordinary light rays, X–rays,$\gamma$–rays, etc. are called electromagnetic radiations because similar waves can be produced by moving a charged body in a magnetic field or a magnet in an electric field. These radiations have wave characteristics and do not require any medium for their propagation.

$\bullet$Wavelength ($\lambda$): The distance between two neighbouring troughs or crests is known as wavelength. It is denoted by $\lambda$ and is expressed in cm, m, nanometers (1 nm =10^–9 m) or Angstrom (1 Å=10^–10 m).

$\bullet$Frequency ($\nu$): The frequency of a wave is the number of times a wave passes through a given point in a medium in one second. It is denoted by $\nu$(nu) and is expressed in cycles per second (cps) or hertz (Hz) 1Hz = 1cps.

The frequency of a wave is inversely proportional to its wave length $\left(\lambda \right)$

$\nu$ $\propto$ $\frac{1}{\lambda }$ or $\nu$ = $\frac{c}{\lambda }$

$\bullet$Velocity: The distance travelled by the wave in one second is called its velocity. It is denoted by c and is expressed in cm sec^–1.

c = $\nu \lambda$ or $\lambda$ = $\frac{c}{\nu }$

$\bullet$Wave number$\left(\bar{\nu }\right)$: It is defined as number of wavelengths per cm. It is denoted by $\bar{\nu }$and is expressed in cm^–1.

$\bar{\nu }$= $\frac{1}{\lambda }$ or $\bar{\nu }$= $\frac{\nu }{c}$

$\bullet$Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted by a. It determines the intensity or brightness of the beam of light.

$\bullet$Electromagnetic Spectrum: The arrangement of the various types of electromagnetic radiation in order of increasing or decreasing wavelengths or frequencies is known as electromagnetic spectrum.

Wavelengths of Electromagnetic Radiations:

Illustration 5. Find out the longest wavelength of absorption line for hydrogen gas containing atoms in ground state.

Solution: $\frac{1}{\lambda }=R{Z}^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$

For longest wavelength DE should be smallest, i.e. transition occurs from

n = 1 to n = 2

i.e. $\frac{1}{\lambda }$= 109673 cm^–1 x 1² $\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$

$\therefore$ $\frac{1}{\lambda }$ = 109673 x $\frac{3}{4}$ cm^–1

$\therefore \lambda$ = $\frac{4}{3\times 109673c{m}^{-1}}$= 1.2157 x 10^–5 cm = 121.6 nm

ATOMIC SPECTRUM

If the atom gains energy the electron passes from a lower energy level to a higher energy level, energy is absorbed that means a specific wave length is absorbed. Consequently, a dark line will appear in the spectrum. This dark line constitutes the absorption spectrum.

If the atom loses energy, the electron passes from higher to a lower energy level, energy is released and a spectral line of specific wavelength is emitted. This line constitutes the emission spectrum.

Types of Emission Spectra:

$\bullet$Continuous spectra: When white light from any source such as sun or bulb is analyzed by passing through a prism, it splits up into seven different wide bands of colour from violet to red (like rainbow). These colour are so continuous that each of them merges into the next. Hence the spectrum is called as continuous spectrum.

$\bullet$Line spectra: When an electric discharge is passed through a gas at low pressure light is emitted. If this light is resolved by a spectroscope, It is found that some isolated coloured lines are obtained on a photographic plate separated from each other by dark spaces. This spectrum is called line spectrum. Each line in the spectrum corresponds to a particular wavelength. Each element gives its own characteristic spectrum.

PLANCK'S QUANTUM THEORY

When a black body is heated, it emits thermal radiations of different wavelengths or frequency. To explain these radiations, Max Planck put forward a theory known as Planck's quantum theory. The main points of quantum theory are

$\bullet$Substances radiate or absorb energy discontinuously in the form of small packets or bundles of energy.

$\bullet$The smallest packet of energy is called quantum. In case of light the quantum is known as photon.

$\bullet$The energy of a quantum is directly proportional to the frequency of the radiation. E $\propto \nu$ (or) E = h$\nu$ where $\nu$ is the frequency of radiation and h is Planck's constant having the value 6.626 x 10^–27 erg – sec or 6.626 x 10^–34 J–sec.

$\bullet$A body can radiate or absorb energy in whole number multiples of a

quantum h$\nu$, 2h$\nu$,3h$\nu$………..nh$\nu$. where 'n' is the positive integer.

$\bullet$Neils Bohr used this theory to explain the structure of atom.

Illustration 9. The wave number of a radiation is 400 cm^-1. Find out its

(a) Wavelength (b) Frequency

(c) J per photon (d) kcal per mol of photons

(e) kJ per mol of photons

Solution: (a) or $\lambda$ = $\frac{1}{\bar{\nu }}$ = $\frac{1}{400c{m}^{-1}}$ = 2.5 x 10^–3 cm

(b) n = $\frac{c}{\lambda }$ = $c\times \bar{\nu }$ = 3 x 10^–10 cm/s x 400 cm^–1 = $1.2\times {10}^{-7}{s}^{-1}$

(c) E_photon = h$\nu$ = $\frac{hc}{\lambda }$ = $h\times c\times \bar{\nu }$

= 6.626 x 10^–34 Js x 3 x 10⁸ cm/s x 400 cm^–1

= 7.95 x 10^–21J

(d) E_photon = 7.95 x 10^–21J

For 1 mol of photon energy = 7.95 x 10^–21J x 6.022 x 10²³ mol^–1

= (4.7875 x 10³ J mol^–1) (1 kcal/4184 J)

= 1.14 kcal mol^–1

(e) E = (4.7875 x 10³ J mol^–1) (1 kJ/1000J)

= 4.7875 kJ mol^–1

^