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November 22, 2024Research into the electric discharge in the discharge tube (J.J. Thomson, 1896) led to electron discovery. The discharge tube contains a glass tube with metal electrodes fused in the walls(Fig.). A pump can be used to draw air through a glass side-arm. The electrodes are connected to a \(10,000\)-volt source, and the air is partially evacuated. The leftover gas in the tube begins to glow as the electric discharge flows between the electrodes. The glow is replaced by faintly bright ‘rays’ that cause fluorescence on the glass at the end distant from the cathode when almost all gas is evacuated from within the tube. Cathode rays originate from the cathode and go away from it in straight lines at right angles.
The electric and magnetic field influence on cathode rays is counterbalanced. Thomson was able to calculate the cathode particle’s charge to mass ratio (e/m). The e/m of cathode particles is – \(1.76 \times {10^8}\) coulombs per gram in SI units. Thomson demonstrated that the value of e/m of the cathode particle was the same regardless of the gas or the metal used to make the cathode through several tests. This demonstrated that the particles that make up cathode rays are all the same and are constituent portions of the various atoms. H.A. Lorentz, a Dutch physicist, gave them the name Electrons.
The impact of \({\rm{X}}\)-rays or ultraviolet light on metals, as well as heated filaments, produce electrons. Radioactive compounds emit these as \(\beta \)-particles as well. As a result, electrons are considered a universal element of all atoms.
J.J. Thomson (1897) used the device depicted in the figure to determine an electron’s charge to mass (e/m) ratio. On the fluorescent screen, electrons generate a bright glowing spot at \({\rm{X}}\). The electrons are first deflected in a circular route by a magnetic field, while the spot is relocated to \({\rm{Y}}\). The radius of the circular route can be determined based on the dimensions of the equipment, the current and number of turns in the electromagnet’s coil, and the angle of deflection of the spot. An electromagnetic field of known strength is then used to return the spot to its original position. The velocity of the electrons can then be calculated using the strength of the electrostatic and magnetic fields.
Magnetic force on the electron beam is equivalent to centrifugal force.
\({\rm{Bev = }}\frac{{{\rm{m}}{{\rm{v}}^{\rm{2}}}}}{{{\rm{ra}}}}\)
Where \(B = \) magnetic field strength
\(v = \) velocity of electrons
\(e = \) charge on the electron
\(m = \) mass of the electron
\(r = \) radius of the circular path of the electron in the magnetic field.
This means:
\(\frac{e}{m} = \frac{v}{{Br}}\)
The value of r is calculated using the tube size and the displacement of the electron spot on the fluorescent screen.
When the strengths of the electrostatic and magnetic fields are counterbalanced,
\(Bev = Ee\) ….(1)
Where \({\rm{E}} = \) strength of the electrostatic field.
Thus:
\({\rm{v}} = \frac{{\rm{E}}}{{\rm{B}}}\) …(2)
If \({\rm{E}}\) and \({\rm{B}}\) are known, \({\rm{v}}\) may be computed, and the value of e/m can be obtained by substituting in equation (1).
\(\frac{{\rm{e}}}{{\rm{m}}}{\rm{ = }}\frac{{\rm{E}}}{{{{\rm{B}}^{\rm{2}}}{\rm{r}}}}\)
Experimentally, all of the quantities on the right side of the equation may be determined. Using this method, the \({\rm{e/m}}\) ratio comes out to \(-1.76 \times {10^8}\) coulomb per gram.
or e/m for the electron \( = – 1.76 \times {10^8}\) coulomb/g
R.A. Millikan (1908) used what is known as Millikan’s Oil-drop Experiment to determine the absolute value of an electron’s charge. The device he employed is depicted in Fig. He sprayed oil droplets into the equipment with an atomizer. Through a hole in the upper plate, an oil droplet falls. The air between the plates is then subjected to \({\rm{X}}\)-rays, which cause the air molecules to expel electrons. The oil droplet captures some of these electrons, giving it a negative charge. The droplet falls due to the influence of gravity when the plates are earthed. He regulated the electric field strength between the two charged plates so that a single oil drop remained stationary, not rising or sinking. The upward force arising due to the negative charge on the drop had barely equaled the weight of the drop at this moment. Electrons are created when \({\rm{X}}\)-rays strike air molecules. One or more electrons are captured by the drop, resulting in a negative charge, \({\rm{Q}}\). Thus,
\({\rm{Q = ne}}\)
Where “\({\rm{n}}\)” denotes the number of electrons and \({\rm{e}}\) denotes the electron’s charge. Millikan determined that the electron had a charge of \({\rm{ – 1}}{\rm{.60}} \times {\rm{1}}{{\rm{0}}^{ – 19}}\) coulombs based on measurements with various drops.
The mass of an electron can be determined using Thomson’s value of \({\rm{e/m}}\) and Millikan’s value of \({\rm{e}}\).
\({\rm{e/m}} = – 1.76 \times {10^8}{\rm{coulomb/g}}\)
\({\rm{e}} = – 1.60 \times {10^{ – 19}}{\rm{coulomb}}\)
\(\frac{{\rm{e}}}{{{\rm{e/m}}}}{\rm{ = }}\frac{{{\rm{1}}{\rm{.60 \times 1}}{{\rm{0}}^{{\rm{ – 19}}}}}}{{{\rm{1}}{\rm{.76 \times 1}}{{\rm{0}}^{\rm{8}}}}}\)
\({\rm{m = 9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ – 28}}}}{\rm{g}}\) or \({\rm{9}}{\rm{.1 \times 1}}{{\rm{0}}^{{\rm{ – 31}}}}{\rm{kg}}\)
The mass of an electron was found to be \(\frac{{\rm{1}}}{{{\rm{1835}}}}{\rm{th}}\) mass of a hydrogen atom.
When the charge and mass of an electron are known, it can be defined as follows: A subatomic particle with a charge of – 1.60 × 10–19 coulomb and a mass of 9.1 x 10–28 g is known as an electron. An electron can also be defined as a particle with a one-unit negative charge and a mass equal to 1/1835th of a hydrogen atom. The charge on an electron was designated as a unit charge by Thomson.
Eugen Goldstein developed a discharge tube with a hole in the cathode in 1886. (Fig.). While cathode rays were moving away from the cathode, he noticed that coloured rays were being produced simultaneously, passing through the perforated cathode and causing a light on the wall opposite the anode. Thomson investigated these rays and discovered that they were made up of particles with a positive charge. He called them positive rays.
When high-speed electrons (cathode rays) collide with a gas molecule in the discharge tube, one or more electrons are knocked out. As a result, a positive ion is produced.
\({\rm{M + }}{{\rm{e}}^{\rm{ – }}} \to {{\rm{M}}^{\rm{ + }}}{\rm{ – 2}}{{\rm{e}}^{\rm{ – }}}\)
Positive rays are formed when positive ions flow through the perforated cathode. When a high-voltage electric discharge is delivered through a gas, the molecules are split into atoms, and the positive atoms (ions) form positive rays.
Thomson and Aston (1913) inferred from a study of the properties of positive rays that an atom consists of at least two parts:
(a) Electrons; and
(b) A positive residue with which the atom’s mass is related.
Protons were discovered in the discharge tube containing hydrogen by E. Goldstein (1886)
\({\rm{H}} \to \mathop {{{\rm{H}}^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ – }}}}\limits_{{\rm{proton}}} \)
J.J. Thomson was the one who investigated their origins. He demonstrated that:
(1) The proton’s real mass is 1.672 × 10– 24 grams. On a relative scale, one atomic mass unit equals one proton (amu).
(2) The electrical charge of the proton is equivalent to that of the electron but has a positive sign. As a result, a proton has a charge of 1.60 x 10–19 coulombs, or + 1 elementary charge unit.
The proton was assumed to be a unit present in all other atoms since it was the lightest positive particle observed in atomic beams in the discharge tube. Protons were also discovered in a number of nuclear reactions, implying that protons are found in all atoms.
As a result, a proton is a subatomic particle with a mass of 1 amu and a charge of 1 elementary charge unit. A proton is a subatomic particle with one unit mass and a positive charge of one unit.
The third subatomic particle was identified in 1932 by Sir James Chadwick. He fired a stream of alpha particles at a target made of beryllium. It was found that a new particle had been discharged. It has roughly the same mass as a proton (1.674 10–24 g) and has no charge.
He gave it the name neutron. A neutron’s assigned relative mass is about one atomic mass unit (amu). As a result, a neutron is a non-charged subatomic particle with a mass nearly equal to that of a proton. A beryllium atom is transformed into a carbon atom through the nuclear reaction in Chadwick’s experiment, which is an example of artificial transmutation.
\({}2^4{\rm{He + }}{}{\rm{4}}^{\rm{9}}{\rm{Be}} \to {}{\rm{6}}^{{\rm{12}}}{\rm{C + }}{}{\rm{0}}^{\rm{1}}{\rm{n}}\)
1. There have been a huge number of subatomic particles found. The three most important particles for us are the proton, neutron, and electron.
2. A subatomic particle with a charge of – 1.60 × 10–19 coulomb and a mass of 9.1 x 10–28 g is known as an electron.
3. The proton’s real mass is 1.672 × 10– 24 grams. On a relative scale, one atomic mass unit equals one proton (amu).
4. The electrical charge of the proton is equivalent to that of the electron but has a positive sign. As a result, a proton has a charge of 1.60 x 10–19 coulombs, or + 1 elementary charge unit.
5. Neutron has roughly the same mass as a proton (1.674 10–24 g) and has no charge.
Q.1. What are the fundamental particles?
Ans: According to John Dalton (1805), all matter comprises minuscule particles known as atoms. He envisioned the atom as a single solid, hard particle that could not be separated. By the end of the nineteenth century, scientists had accumulated enough experimental evidence to conclude that the atom comprises even smaller particles. The fundamental particles are the subatomic particles that make up the universe. There have been a huge number of subatomic particles found. The three most important particles for us are the proton, neutron, and electron.
Q.2. Who discovered the electron?
Ans: The electron was discovered due to research into the electric discharge in the discharge tube by J.J. Thomson in1896.
Q.3. What is the relationship between the mass of an electron and the mass of a hydrogen atom?
Ans: The mass of an electron was found to be \(\frac{1}{{1835}}{\rm{th}}\) of the mass of a hydrogen atom.
Q.4. What is the mass and charge of a proton?
Ans: The proton’s real mass is 1.672 × 10– 24 grams. On a relative scale, one atomic mass unit equals one proton (amu). The electrical charge of the proton is equivalent to that of the electron but has a positive sign. As a result, a proton has a charge of 1.60 x 10–19 coulombs, or + 1 elementary charge unit.