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Structure of Atom - Discovery of Sub-atomic Particles

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Cathode Ray Experiment: Conducted by J.J. Thomson using a discharge tube at very low pressure (10−210^{-2} to 10−3 mm Hg10^{-3} \text{ mm Hg}) and high voltage (10,000 V10,000 \text{ V}). Cathode rays travel from the negative electrode (cathode) to the positive electrode (anode).

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Properties of Cathode Rays: They travel in straight lines, possess kinetic energy, and are deflected by electric and magnetic fields toward the positive plate, indicating they consist of negatively charged particles called electrons (e−e^-).

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Charge-to-Mass Ratio (e/me/m): J.J. Thomson determined the e/me/m ratio of an electron to be 1.758820×1011 C kg−11.758820 \times 10^{11} \text{ C kg}^{-1}. This value is independent of the nature of the gas or the electrode material used.

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Charge of Electron: R.A. Millikan determined the charge on an electron using the Oil Drop Experiment. The charge (ee) was found to be −1.6022×10−19 C-1.6022 \times 10^{-19} \text{ C}.

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Discovery of Protons: E. Goldstein observed 'Canal Rays' or 'Anode Rays' in a modified discharge tube. Unlike cathode rays, the properties of anode rays depend on the nature of the gas present in the tube. The smallest and lightest positive ion was obtained from hydrogen and called a proton (p+p^+).

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Discovery of Neutrons: James Chadwick (1932) discovered neutrons by bombarding a thin sheet of beryllium with α\alpha-particles. Neutrons are electrically neutral particles with a mass slightly greater than that of a proton.

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Sub-atomic Particle Properties: Electron mass ≈9.1×10−31 kg\approx 9.1 \times 10^{-31} \text{ kg}, Proton mass ≈1.672×10−27 kg\approx 1.672 \times 10^{-27} \text{ kg}, and Neutron mass ≈1.675×10−27 kg\approx 1.675 \times 10^{-27} \text{ kg}.

📐Formulae

me=ee/mm_e = \frac{e}{e/m}

Q=ne (Quantization of charge)Q = ne \text{ (Quantization of charge)}

24He+49Be→612C+01n^{4}_{2}\text{He} + ^{9}_{4}\text{Be} \rightarrow ^{12}_{6}\text{C} + ^{1}_{0}\text{n}

Mass of electron in amu≈0.00054 u\text{Mass of electron in amu} \approx 0.00054 \text{ u}

Mass of proton in amu≈1.00727 u\text{Mass of proton in amu} \approx 1.00727 \text{ u}

Mass of neutron in amu≈1.00867 u\text{Mass of neutron in amu} \approx 1.00867 \text{ u}

💡Examples

Problem 1:

Calculate the mass of one mole of electrons.

Solution:

Mass of one electron ≈9.109×10−31 kg\approx 9.109 \times 10^{-31} \text{ kg}. Number of electrons in one mole (Avogadro's number) NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23} \text{ mol}^{-1}. Total mass=(9.109×10−31 kg)×(6.022×1023)≈5.48×10−7 kg=0.548 mg\text{Total mass} = (9.109 \times 10^{-31} \text{ kg}) \times (6.022 \times 10^{23}) \approx 5.48 \times 10^{-7} \text{ kg} = 0.548 \text{ mg}.

Explanation:

To find the mass of a mole of any subatomic particle, multiply the mass of a single particle by Avogadro's constant (NAN_A).

Problem 2:

In a Millikan's oil drop experiment, the static electric charge on the drops was found to be −3.2×10−19 C-3.2 \times 10^{-19} \text{ C} and −4.8×10−19 C-4.8 \times 10^{-19} \text{ C}. What is the smallest possible magnitude of charge according to these observations?

Solution:

Charge Q1=−3.2×10−19 CQ_1 = -3.2 \times 10^{-19} \text{ C}, Q2=−4.8×10−19 CQ_2 = -4.8 \times 10^{-19} \text{ C}. The charges must be integral multiples of a basic unit ee. Q1Q2=−3.2−4.8=23\frac{Q_1}{Q_2} = \frac{-3.2}{-4.8} = \frac{2}{3}. The Highest Common Factor (HCF) of the magnitudes is 1.6×10−19 C1.6 \times 10^{-19} \text{ C}.

Explanation:

Millikan found that charge is quantized (Q=neQ = ne), meaning all observed charges are integer multiples of the fundamental electronic charge e=1.602×10−19 Ce = 1.602 \times 10^{-19} \text{ C}.