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Structure of Atom - Discovery of Protons-advanced

Grade 9CBSE

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

🔑Concepts

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Discovery: In 18861886, E.GoldsteinE. Goldstein discovered the presence of new radiations in a gas discharge tube and called them Canal Rays (or Anode Rays).

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Experimental Setup: A gas discharge tube was used with a perforated cathode. When a high voltage was applied, radiations were observed passing through the holes (canals) of the cathode.

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Nature of Canal Rays: These rays consist of positively charged particles. Unlike cathode rays, the nature of these rays depends on the gas present in the discharge tube.

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Origin: Canal rays are not emitted from the anode. They are formed when high-speed cathode rays (electrons) collide with the neutral gas atoms in the tube, knocking out electrons and leaving behind positively charged ions.

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Charge-to-Mass Ratio (e/me/m): The e/me/m ratio for canal rays is not constant and varies with the nature of the gas used in the tube. It is maximum when Hydrogen gas is used.

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The Proton: The lightest and smallest positive ion was obtained from Hydrogen. This particle was named the Proton.

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Mass of Proton: The mass of a proton is approximately 1.672×10−27 kg1.672 \times 10^{-27} \text{ kg}, which is about 18361836 times the mass of an electron.

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Charge of Proton: The charge of a proton is equal in magnitude but opposite in sign to that of an electron: +1.602×10−19 C+1.602 \times 10^{-19} \text{ C}.

📐Formulae

mp≈1836×mem_p \approx 1836 \times m_e

Charge of proton (qp)=+1.6×10−19 C\text{Charge of proton } (q_p) = +1.6 \times 10^{-19} \text{ C}

Mass of proton (mp)=1.672×10−27 kg≈1 u\text{Mass of proton } (m_p) = 1.672 \times 10^{-27} \text{ kg} \approx 1 \text{ u}

Net Charge of Atom=(Z×(+e))+(Z×(−e))=0\text{Net Charge of Atom} = (Z \times (+e)) + (Z \times (-e)) = 0

💡Examples

Problem 1:

If the mass of an electron is 9.1×10−31 kg9.1 \times 10^{-31} \text{ kg} and the mass of a proton is 1.67×10−27 kg1.67 \times 10^{-27} \text{ kg}, calculate the ratio of the mass of a proton to the mass of an electron.

Solution:

Ratio=mpme=1.67×10−27 kg9.1×10−31 kg\text{Ratio} = \frac{m_p}{m_e} = \frac{1.67 \times 10^{-27} \text{ kg}}{9.1 \times 10^{-31} \text{ kg}} Ratio≈1835.16\text{Ratio} \approx 1835.16

Explanation:

The mass of a proton is approximately 18361836 times heavier than that of an electron, which is why the mass of an electron is considered negligible when calculating the atomic mass.

Problem 2:

Explain why the e/me/m ratio of canal rays depends on the gas in the discharge tube, whereas for cathode rays it is independent.

Solution:

em (Canal Rays)∝1Mass of Gas Ion\frac{e}{m} \text{ (Canal Rays)} \propto \frac{1}{\text{Mass of Gas Ion}} em (Cathode Rays)=Constant (for all gases)\frac{e}{m} \text{ (Cathode Rays)} = \text{Constant (for all gases)}

Explanation:

Cathode rays consist of electrons, which are fundamental particles identical for all matter. However, canal rays consist of ionized gas atoms. Since different gases have different atomic masses, their charge-to-mass ratio varies.

Problem 3:

An atom has a nucleus containing 1212 protons. What is the total positive charge in Coulombs?

Solution:

Total Charge=n×qp\text{Total Charge} = n \times q_p Total Charge=12×(+1.6×10−19 C)\text{Total Charge} = 12 \times (+1.6 \times 10^{-19} \text{ C}) Total Charge=+19.2×10−19 C=+1.92×10−18 C\text{Total Charge} = +19.2 \times 10^{-19} \text{ C} = +1.92 \times 10^{-18} \text{ C}

Explanation:

The total positive charge of a nucleus is the product of the number of protons and the elementary charge of a single proton.