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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 10−3extmmHg10^{-3} ext{ mm Hg}) and high voltage (10,000extV10,000 ext{ 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 1.758820imes1011extCkg−11.758820 imes 10^{11} ext{ 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 −1.6022imes10−19extC-1.6022 imes 10^{-19} ext{ 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 ≈9.1imes10−31extkg\approx 9.1 imes 10^{-31} ext{ kg}, Proton mass ≈1.672imes10−27extkg\approx 1.672 imes 10^{-27} ext{ kg}, and Neutron mass ≈1.675imes10−27extkg\approx 1.675 imes 10^{-27} ext{ 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 ≈9.109imes10−31extkg\approx 9.109 imes 10^{-31} ext{ kg}. Number of electrons in one mole (Avogadro's number) NA=6.022imes1023extmol−1N_A = 6.022 imes 10^{23} ext{ mol}^{-1}. Total mass=(9.109imes10−31extkg)imes(6.022imes1023)≈5.48imes10−7extkg=0.548extmg\text{Total mass} = (9.109 imes 10^{-31} ext{ kg}) imes (6.022 imes 10^{23}) \approx 5.48 imes 10^{-7} ext{ kg} = 0.548 ext{ 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 −3.2imes10−19extC-3.2 imes 10^{-19} ext{ C} and −4.8imes10−19extC-4.8 imes 10^{-19} ext{ C}. What is the smallest possible magnitude of charge according to these observations?

Solution:

Charge Q1=−3.2imes10−19extCQ_1 = -3.2 imes 10^{-19} ext{ C}, Q2=−4.8imes10−19extCQ_2 = -4.8 imes 10^{-19} ext{ 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 1.6imes10−19extC1.6 imes 10^{-19} ext{ C}.

Explanation:

Millikan found that charge is quantized (Q=neQ = ne), meaning all observed charges are integer multiples of the fundamental electronic charge e=1.602imes10−19extCe = 1.602 imes 10^{-19} ext{ C}.

Discovery of Sub-atomic Particles Revision - Class 11 Chemistry CBSE