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Structure of Atom - Discovery of Subatomic Particles-advanced

Grade 9CBSE

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

🔑Concepts

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Dalton's Atomic Theory suggested atoms were indivisible, but the discovery of subatomic particles like electrons, protons, and neutrons proved otherwise.

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The electron was discovered by J.J. Thomson using Cathode Ray Tube (CRT) experiments. Cathode rays are streams of fast-moving electrons that travel from the cathode (negative electrode) to the anode (positive electrode).

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The charge of an electron is approximately −1.602×10−19 C-1.602 \times 10^{-19} \text{ C}, and its mass is 9.1×10−31 kg9.1 \times 10^{-31} \text{ kg}, which is about 11840\frac{1}{1840} of the mass of a hydrogen atom.

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Protons were discovered by E. Goldstein through 'Canal Rays' (Anode Rays) experiments. These rays consist of positively charged particles. The mass of a proton is approximately 1.672×10−27 kg1.672 \times 10^{-27} \text{ kg}.

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Neutrons were discovered by James Chadwick in 19321932 by bombarding a thin sheet of beryllium with α\alpha-particles. Neutrons are electrically neutral and have a mass slightly greater than that of a proton (1.675×10−27 kg1.675 \times 10^{-27} \text{ kg}).

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The specific charge (e/me/m ratio) of an electron is a constant value of 1.76×1011 C/kg1.76 \times 10^{11} \text{ C/kg}, regardless of the gas used in the discharge tube.

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Rutherford's Alpha-particle scattering experiment led to the discovery of the atomic nucleus. He concluded that most of the atom is empty space and the mass is concentrated in a tiny, positively charged center called the nucleus.

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The size of the nucleus is approximately 10−15 m10^{-15} \text{ m}, while the size of the atom is approximately 10−10 m10^{-10} \text{ m}. Thus, the radius of an atom is about 10510^5 times the radius of the nucleus.

📐Formulae

Mass of electron (me)≈9.1×10−31 kg\text{Mass of electron } (m_e) \approx 9.1 \times 10^{-31} \text{ kg}

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

Mass of neutron (mn)≈1.675×10−27 kg\text{Mass of neutron } (m_n) \approx 1.675 \times 10^{-27} \text{ kg}

Charge of electron (e)=−1.602×10−19 C\text{Charge of electron } (e) = -1.602 \times 10^{-19} \text{ C}

Mass Number (A)=Number of protons (Z)+Number of neutrons (n)\text{Mass Number } (A) = \text{Number of protons } (Z) + \text{Number of neutrons } (n)

💡Examples

Problem 1:

An ion X2+X^{2+} contains 1010 electrons and 1212 neutrons. Determine the atomic number (ZZ) and the mass number (AA) of the element XX.

Solution:

In an ion X2+X^{2+}, the number of electrons is 1010. Since it has a +2+2 charge, it means the neutral atom lost 22 electrons. Number of electrons in neutral atom =10+2=12= 10 + 2 = 12. Atomic Number (ZZ) = Number of protons = Number of electrons in neutral atom =12= 12. Mass Number (AA) = Protons + Neutrons =12+12=24= 12 + 12 = 24.

Explanation:

Atomic number is defined by the proton count, which remains unchanged during ion formation. The mass number is the sum of protons and neutrons.

Problem 2:

Calculate the ratio of the mass of a proton to the mass of an electron.

Solution:

Mass of proton mp≈1.672×10−27 kgm_p \approx 1.672 \times 10^{-27} \text{ kg}. Mass of electron me≈9.1×10−31 kgm_e \approx 9.1 \times 10^{-31} \text{ kg}. Ratio=1.672×10−279.1×10−31≈1837\text{Ratio} = \frac{1.672 \times 10^{-27}}{9.1 \times 10^{-31}} \approx 1837

Explanation:

This shows that the mass of an electron is negligible (nearly 1/18401/1840 times) compared to the mass of a proton.

Problem 3:

During the discovery of neutrons, beryllium was bombarded with alpha particles. Complete the nuclear equation: 49Be+24He→??C+01n^9_4\text{Be} + ^4_2\text{He} \rightarrow ^?_?\text{C} + ^1_0\text{n}

Solution:

According to the conservation of mass number: 9+4=A+1  ⟹  A=129 + 4 = A + 1 \implies A = 12. According to the conservation of atomic number: 4+2=Z+0  ⟹  Z=64 + 2 = Z + 0 \implies Z = 6. The completed equation is: 49Be+24He→612C+01n^9_4\text{Be} + ^4_2\text{He} \rightarrow ^{12}_6\text{C} + ^1_0\text{n}

Explanation:

In nuclear reactions, the sum of mass numbers and the sum of atomic numbers must be balanced on both sides of the equation.