Discovery of Sub-atomic Particles

Introduction to Sub-Atomic Particles

For a long time, atoms were considered indivisible, as proposed by John Dalton. However, experiments conducted towards the end of the 19th and the beginning of the 20th century turned this idea upside down. Scientists discovered that atoms are actually made up of smaller, sub-atomic particles: electrons, protons, and neutrons.

Before diving into the experiments, keep this fundamental rule of electrostatics in mind:

Key Point: "Like charges repel each other and unlike charges attract each other."

The Particulate Nature of Electricity

In 1830, Michael Faraday showed that passing electricity through an electrolyte solution caused chemical reactions at the electrodes, leading to the liberation and deposition of matter. This was the first major hint that electricity has a particulate nature.

Discovery of Electron

In the mid-1850s, scientists (mainly Faraday) began studying electrical discharge in partially evacuated glass tubes, known as cathode ray discharge tubes.

The Cathode Ray Tube Experiment

A cathode ray tube consists of a glass tube containing two thin pieces of metal (electrodes) sealed inside.

Experimental Conditions:

1. Very low pressure: Achieved by evacuating the glass tube using a vacuum pump.
2. Very high voltage: Applied across the electrodes.

Under these conditions, current starts flowing as a stream of particles moving from the negative electrode (cathode) to the positive electrode (anode). These streams are called cathode rays or cathode ray particles.

To confirm this flow, scientists made a hole in the anode and coated the glass tube behind it with a phosphorescent material (zinc sulphide). When the rays passed through the hole and struck the coating, a bright spot developed.

Characteristics of Cathode Rays

1. They start from the cathode and move towards the anode.
2. They are invisible themselves, but their behaviour can be observed using fluorescent or phosphorescent materials (e.g., television picture tubes are cathode ray tubes coated with fluorescent materials).
3. In the absence of electrical or magnetic fields, they travel in straight lines.
4. In the presence of electrical or magnetic fields, they deflect towards the positive plate, behaving like negatively charged particles. These particles were named electrons.
5. [School Exam Focus] The characteristics of cathode rays (electrons) do not depend upon the material of the electrodes or the nature of the gas present in the tube.

Conclusion: Electrons are a basic, universal constituent of all atoms.

Charge to Mass Ratio of Electron

In 1897, British physicist J.J. Thomson measured the ratio of electrical charge ($e$) to the mass of an electron ($m_e$).

He used a modified cathode ray tube and applied electrical and magnetic fields perpendicular to each other, as well as to the path of the electrons.

Thomson's Observations

• When only the electric field is applied, electrons deviate and hit point A on the screen.
• When only the magnetic field is applied, electrons strike point C.
• By carefully balancing both fields, the electrons pass straight through (as if no field were present) and hit point B.

Factors Affecting Deflection

Thomson argued that the amount of deviation depends on three critical factors:

1. Magnitude of the negative charge: Greater the charge on the particle, greater the interaction with the fields, resulting in greater deflection.
2. Mass of the particle: Lighter particles experience greater deflection.
3. Strength of the field: Deflection increases with an increase in the voltage across the electrodes or the strength of the magnetic field.

[JEE Tip] Deflection $\propto \frac{e}{m_e} \times \text{Field Strength}$

The $e/m_e$ Value

By carrying out accurate measurements, Thomson determined the charge-to-mass ratio:

$$\frac{e}{m_e} = 1.758820 \times 10^{11} \text{ C kg}^{-1}$$

Where:

• $m_e$ = mass of the electron in kg
• $e$ = magnitude of the charge on the electron in coulomb (C) (Note: Since electrons are negatively charged, the actual charge is $-e$.)

Charge on the Electron

While Thomson found the ratio, it was R.A. Millikan (1906-14) who determined the exact charge of an electron using his famous Oil Drop Experiment.

Millikan found the charge on the electron to be $-1.6 \times 10^{-19} \text{ C}$. The present accepted value is: Charge of an electron ($e$) = $-1.602176 \times 10^{-19} \text{ C}$

Calculating the Mass of an Electron

By combining Millikan's charge value with Thomson's $e/m_e$ ratio, the exact mass of the electron ($m_e$) was calculated:

$$m_e = \frac{e}{e/m_e}$$

$$m_e = \frac{1.602176 \times 10^{-19} \text{ C}}{1.758820 \times 10^{11} \text{ C kg}^{-1}}$$

$$m_e = 9.1094 \times 10^{-31} \text{ kg}$$

[NEET Important] You must memorize these three values: the charge of an electron, the $e/m_e$ ratio, and the mass of the electron. They are frequently used in numerical problems across Physical Chemistry and Modern Physics.

Discovery of Protons and Neutrons

Discovery of Protons (Canal Rays)

Electrical discharge carried out in a modified cathode ray tube led to the discovery of positively charged particles called canal rays.

Characteristics of Canal Rays:

1. [Exception Alert] Unlike cathode rays, the mass of positively charged particles depends upon the nature of the gas present in the tube. They are simply positively charged gaseous ions.
2. The charge-to-mass ratio ($e/m$) of these particles depends on the gas from which they originate.
3. Some of these positively charged particles carry a multiple of the fundamental unit of electrical charge.
4. Their behaviour in magnetic or electrical fields is opposite to that of electrons (they deflect towards the negative plate).

The smallest and lightest positive ion was obtained from hydrogen and was called a proton. It was characterised in 1919.

Discovery of Neutrons

Later, scientists realized that atoms must contain an electrically neutral particle to account for their total mass.

In 1932, Chadwick discovered these particles by bombarding a thin sheet of beryllium with $\alpha$-particles.

• Electrically neutral particles were emitted.
• Their mass was found to be slightly greater than that of protons.
• He named these particles neutrons.

Memory Capsule

Quick Revision Summary

• Faraday (1830): Particulate nature of electricity (Electrolysis).
• Cathode Rays: Stream of electrons; independent of gas/electrode; travel cathode $\rightarrow$ anode.
• J.J. Thomson (1897): Discovered $e/m_e$ ratio using perpendicular electric and magnetic fields.
• R.A. Millikan (1906-14): Discovered charge of electron via Oil Drop Experiment.
• Canal Rays: Positively charged gaseous ions; dependent on the gas used.
• Proton: Lightest positive ion (from Hydrogen), characterised in 1919.
• Chadwick (1932): Discovered neutron by bombarding Beryllium with $\alpha$-particles.

Crucial Constants Formula Sheet

Property	Symbol	Value
Charge of Electron	$e$	$-1.602176 \times 10^{-19} \text{ C}$
Charge-to-Mass Ratio	$e/m_e$	$1.758820 \times 10^{11} \text{ C kg}^{-1}$
Mass of Electron	$m_e$	$9.1094 \times 10^{-31} \text{ kg}$

Mnemonic for Discoveries

"PEN - RTC"

• Proton $\rightarrow$ Rutherford (Characterised in 1919, though canal rays were earlier)
• Electron $\rightarrow$ Thomson
• Neutron $\rightarrow$ Chadwick

Example 1: Calculating Mass of Electron from Experimental Data

Using J.J. Thomson's charge-to-mass ratio ($1.758820 \times 10^{11} \text{ C kg}^{-1}$) and Millikan's charge of an electron ($1.602176 \times 10^{-19} \text{ C}$), calculate the mass of an electron in kg.

Solution:

1. Identify the given values:

• Magnitude of charge, $e = 1.602176 \times 10^{-19} \text{ C}$
• Charge-to-mass ratio, $e/m_e = 1.758820 \times 10^{11} \text{ C kg}^{-1}$

1. Set up the formula: We know that: $$m_e = \frac{e}{e/m_e}$$

2. Substitute the values: $$m_e = \frac{1.602176 \times 10^{-19} \text{ C}}{1.758820 \times 10^{11} \text{ C kg}^{-1}}$$

3. Calculate: $$m_e = 0.910938 \times 10^{-30} \text{ kg}$$ $$m_e = 9.1094 \times 10^{-31} \text{ kg}$$

Takeaway: The mass of an electron is incredibly small, on the order of $10^{-31}$ kg. This calculation bridges Thomson's and Millikan's historic experiments.

Example 2: Comparing Deflection of Particles

Two hypothetical particles, A and B, enter a uniform electric field. Particle A has a charge of $-e$ and mass $m$. Particle B has a charge of $-2e$ and mass $m/2$. Which particle will show greater deflection and by what factor? Assume initial velocities are identical.

Solution:

1. Recall Thomson's principle: The amount of deviation (deflection) depends directly on the magnitude of the charge and inversely on the mass. $$\text{Deflection} \propto \frac{\text{Charge}}{\text{Mass}}$$

2. Calculate ratio for Particle A: $$\text{Ratio}_A = \frac{e}{m}$$

3. Calculate ratio for Particle B: $$\text{Ratio}_B = \frac{2e}{m/2} = \frac{4e}{m}$$

4. Compare the ratios: $$\frac{\text{Deflection}_B}{\text{Deflection}_A} = \frac{4e/m}{e/m} = 4$$

Answer: Particle B will show 4 times greater deflection than Particle A.

Takeaway: Deflection is highly sensitive to both charge and mass. Lighter particles with higher charges deflect the most.

Example 3: Multiple Fundamental Charges

During an experiment with canal rays, a positively charged gaseous ion is found to carry a charge that is 3 times the fundamental unit of electrical charge. Calculate the magnitude of the charge on this ion in Coulombs.

Solution:

1. Identify the fundamental unit of charge: The fundamental unit of electrical charge ($e$) is the magnitude of the charge on an electron. $$e = 1.602176 \times 10^{-19} \text{ C}$$

2. Set up the calculation: Charge on the ion ($q$) = $3 \times e$

3. Substitute and solve: $$q = 3 \times 1.602176 \times 10^{-19} \text{ C}$$ $$q = 4.806528 \times 10^{-19} \text{ C}$$

Answer: The magnitude of the charge on the ion is $4.806528 \times 10^{-19} \text{ C}$.

Takeaway: As stated in the NCERT text, positively charged particles in canal rays can carry multiples of the fundamental unit of electrical charge.

Example 4: Total Mass of a Collection of Electrons

Calculate the total mass of $10^{20}$ electrons. (Given: $m_e = 9.1094 \times 10^{-31} \text{ kg}$)

Solution:

1. Identify given values:

• Mass of one electron ($m_e$) = $9.1094 \times 10^{-31} \text{ kg}$
• Number of electrons ($n$) = $10^{20}$

1. Formula: $$\text{Total Mass} = n \times m_e$$

2. Calculate: $$\text{Total Mass} = 10^{20} \times (9.1094 \times 10^{-31} \text{ kg})$$ $$\text{Total Mass} = 9.1094 \times 10^{-11} \text{ kg}$$

Answer: The total mass of $10^{20}$ electrons is $9.1094 \times 10^{-11} \text{ kg}$.

Takeaway: Even a massive number of electrons ($10^{20}$) contributes a negligible amount of mass to macroscopic objects.

Example 5: Total Charge of a Collection of Electrons

Determine the total charge carried by $5 \times 10^{15}$ electrons.

Solution:

1. Identify given values:

• Charge of one electron = $-1.602 \times 10^{-19} \text{ C}$ (using standard approximation)
• Number of electrons ($n$) = $5 \times 10^{15}$

1. Formula: $$\text{Total Charge} = n \times \text{Charge of one electron}$$

2. Calculate: $$\text{Total Charge} = (5 \times 10^{15}) \times (-1.602 \times 10^{-19} \text{ C})$$ $$\text{Total Charge} = -8.01 \times 10^{-4} \text{ C}$$

Answer: The total charge is $-8.01 \times 10^{-4} \text{ C}$.

Takeaway: Always remember to include the negative sign when calculating the charge of electrons, unless only the magnitude is asked.

Example 6: Conceptual - Changing the Gas in Discharge Tubes

A scientist performs two experiments. In Experiment 1, she uses a cathode ray tube filled with Helium gas. In Experiment 2, she uses a modified cathode ray tube (to study canal rays) filled with Helium gas. She then replaces Helium with Neon in both tubes. How will the properties of the rays change in both tubes?

Solution:

1. Analyze Experiment 1 (Cathode Rays):

• Cathode rays consist of electrons.
• According to NCERT, the characteristics of cathode rays (electrons) do not depend upon the nature of the gas present in the tube.
• Result: No change in the properties or $e/m$ ratio of the cathode rays.

1. Analyze Experiment 2 (Canal Rays):

• Canal rays consist of positively charged gaseous ions.
• According to NCERT, the mass and the charge-to-mass ratio of positively charged particles depend upon the nature of the gas.
• Result: The $e/m$ ratio and mass of the canal rays will change because Neon ions are heavier than Helium ions.

Takeaway: Electrons are universal constituents of matter, but positive ions are specific to the element they originate from.

Example 7: Balancing Fields in Thomson's Experiment

In J.J. Thomson's experiment, electrons hit point A when only the electric field is applied. What must be done to make the electrons hit point B (the straight-line path) without turning off the electric field?

Solution:

1. Understand the setup:

• Point A is the deflection caused by the electric field.
• Point B is the straight-line path (zero net deflection).
• Point C is the deflection caused by the magnetic field.

1. Identify the required action: To bring the electron back to the straight-line path (Point B), a force equal and opposite to the electric force must be applied.

2. Application: A magnetic field must be applied perpendicular to both the electric field and the path of the electrons. The strength of this magnetic field must be carefully balanced so that the magnetic deflection exactly cancels out the electric deflection.

Takeaway: The straight-line path in Thomson's experiment is achieved by balancing two opposing forces: the electric force and the magnetic force.

Example 8: Identifying the Neutral Particle

In 1932, a scientist bombarded a thin sheet of beryllium with $\alpha$-particles and observed the emission of electrically neutral particles with a mass slightly greater than that of protons. Identify the scientist and the particle discovered.

Solution:

1. Recall the historical experiments from the text:

• Faraday (1830): Electrolysis.
• Thomson (1897): $e/m_e$ ratio of electron.
• Millikan (1906-14): Charge of electron.
• Chadwick (1932): Bombardment of Beryllium with $\alpha$-particles.

1. Identify the particle: The electrically neutral particle with a mass slightly greater than a proton is the neutron.

Answer: The scientist was Chadwick, and the particle discovered was the neutron.

Takeaway: Memorize the specific reagents used in Chadwick's experiment: Beryllium target and $\alpha$-particle projectiles.

Example 9: Deflection Dependence on Voltage

In a cathode ray tube, the voltage across the electrodes is doubled. According to Thomson's arguments, how will this affect the deflection of the electrons from their original path?

Solution:

1. Recall Thomson's arguments for deflection: The deflection of electrons from their original path depends on:

• Magnitude of negative charge.
• Mass of the particle.
• Strength of the electrical or magnetic field.

1. Analyze the change: The text states: "the deflection of electrons from its original path increases with the increase in the voltage across the electrodes."

2. Conclusion: By doubling the voltage, the strength of the electrical field increases, which will result in an increase in the deflection of the electrons.

Takeaway: Field strength (voltage) is directly proportional to the amount of deflection.

Example 10: Cathode Ray Tube Conditions

Why are very low pressures required to observe electrical discharge in cathode ray tubes?

Solution:

1. Understand the nature of gases: Under normal atmospheric pressure, gases are poor conductors of electricity.

2. Effect of low pressure: By evacuating the glass tube to a very low pressure, the number of gas molecules inside the tube is drastically reduced.

3. Result: This allows the electrons (cathode rays) emitted from the cathode to travel towards the anode without colliding frequently with gas molecules, enabling the flow of current and the observation of the discharge.

Takeaway: The two critical conditions for cathode ray experiments are very low pressure and very high voltage.