Case Study 1: Discovery of Electrons
In 1897, J.J. Thomson performed experiments with cathode ray tubes. He found that when high voltage was applied across electrodes in a discharge tube, rays travelled from the cathode to the anode. These rays were deflected by electric and magnetic fields, showing that they carried negative charge. Thus, he discovered electrons, the first subatomic particle.
Questions:
- Cathode rays are made up of:
(a) protons (b) neutrons (c) electrons (d) atoms
Answer: (c) - Charge on an electron = ?
(a) +1.6 × 10⁻¹⁹ C (b) –1.6 × 10⁻¹⁹ C (c) 0 (d) +1 C
Answer: (b) - Electron was discovered by:
(a) Rutherford (b) Thomson (c) Millikan (d) Bohr
Answer: (b) - Mass of electron ≈ ?
(a) 9.1 × 10⁻³¹ kg (b) 1.67 × 10⁻²⁷ kg (c) 1 u (d) 0
Answer: (a) - Rays travelling from cathode to anode are called:
(a) canal rays (b) cathode rays (c) X-rays (d) cosmic rays
Answer: (b)
Case Study 2: Discovery of Protons
E. Goldstein discovered positive rays while working with discharge tubes. These rays travelled in the opposite direction of cathode rays and were later called canal rays. They showed the presence of positively charged particles called protons.
Questions:
- Canal rays are made up of:
(a) electrons (b) neutrons (c) protons (d) atoms
Answer: (c) - Who discovered canal rays?
(a) Thomson (b) Goldstein (c) Rutherford (d) Bohr
Answer: (b) - Relative charge on a proton = ?
(a) 0 (b) +1 (c) –1 (d) +2
Answer: (b) - Relative mass of proton = ?
(a) 0 (b) 1/1837 (c) 1 (d) 1837
Answer: (c) - Proton was discovered after:
(a) neutron (b) electron (c) photon (d) atom
Answer: (b)
Case Study 3: Discovery of Neutrons
In 1932, James Chadwick discovered neutrons while studying radiation. Neutrons are neutral particles present in the nucleus of atoms. Their presence explained the difference between atomic number and mass number.
Questions:
- Neutrons were discovered by:
(a) Rutherford (b) Chadwick (c) Bohr (d) Goldstein
Answer: (b) - Neutron has:
(a) positive charge (b) negative charge (c) no charge (d) double charge
Answer: (c) - Mass of neutron ≈ ?
(a) nearly equal to electron (b) nearly equal to proton (c) zero (d) 1/1837 of proton
Answer: (b) - Neutrons are present in:
(a) outside nucleus (b) inside nucleus (c) orbitals (d) none
Answer: (b) - Presence of neutrons explains:
(a) atomic number (b) isotopes (c) mass number (d) ions
Answer: (c)
Case Study 4: Rutherford’s Experiment
In 1909, Rutherford bombarded thin gold foil with alpha particles. Most passed through, some deflected, and very few bounced back. He concluded that the atom has a small, dense, positively charged nucleus where most mass is concentrated, and electrons revolve around it.
Questions:
- Most alpha particles passed through because:
(a) atom is hollow (b) atom has empty space (c) electrons stopped them (d) none
Answer: (b) - Deflection of few alpha particles proved:
(a) electrons exist (b) nucleus exists (c) neutrons exist (d) atoms are indivisible
Answer: (b) - Rutherford’s model failed to explain:
(a) stability of atom (b) existence of protons (c) nucleus (d) electrons
Answer: (a) - Which particle was bombarded on gold foil?
(a) beta particles (b) electrons (c) alpha particles (d) protons
Answer: (c) - Nucleus is:
(a) small, light, negative (b) small, dense, positive (c) large, hollow, positive (d) none
Answer: (b)
Case Study 5: Bohr’s Model
Niels Bohr improved Rutherford’s model. He suggested that electrons revolve in fixed orbits with fixed energy. As long as electrons stay in these shells, they do not lose energy. Electrons can absorb or release energy only by jumping between shells.
Questions:
- Bohr’s model explained:
(a) stability of atom (b) splitting of nucleus (c) radioactivity (d) none
Answer: (a) - Electrons in Bohr’s model revolve in:
(a) any random path (b) fixed orbits (c) nucleus (d) none
Answer: (b) - Energy of orbits is:
(a) continuous (b) quantized (c) infinite (d) zero
Answer: (b) - Jump of electron from higher to lower shell results in:
(a) emission of energy (b) absorption of energy (c) no energy change (d) atom breaks
Answer: (a) - Bohr’s model failed for:
(a) H atom (b) He atom (c) multi-electron atoms (d) none
Answer: (c)
Case Study 6: Atomic Number and Mass Number
In an exam, students were asked to define atomic number and mass number. Atomic number (Z) is the number of protons, while mass number (A) is the sum of protons and neutrons. For example, carbon has Z = 6 and A = 12, meaning 6 protons and 6 neutrons.
Questions:
- Atomic number of carbon = ?
(a) 6 (b) 12 (c) 18 (d) 24
Answer: (a) - Mass number of carbon = ?
(a) 6 (b) 12 (c) 18 (d) 24
Answer: (b) - Neutrons in carbon = ?
(a) 0 (b) 6 (c) 12 (d) 18
Answer: (b) - Atomic number is equal to:
(a) protons (b) neutrons (c) electrons (d) both (a) and (c) in neutral atom
Answer: (d) - Mass number = ?
(a) protons only (b) protons + neutrons (c) neutrons only (d) none
Answer: (b)
Case Study 7: Isotopes
Hydrogen has three isotopes: Protium (¹H), Deuterium (²H), and Tritium (³H). All have one proton but different numbers of neutrons. Isotopes have same chemical properties but different physical properties like mass and radioactivity.
Questions:
- Isotopes have same:
(a) atomic number (b) mass number (c) neutrons (d) all
Answer: (a) - Number of neutrons in Tritium = ?
(a) 0 (b) 1 (c) 2 (d) 3
Answer: (c) - Isotopes differ in:
(a) chemical properties (b) physical properties (c) both (d) none
Answer: (b) - Isotopes are used in:
(a) medicine (b) agriculture (c) industry (d) all
Answer: (d) - Isotope of hydrogen used in nuclear fusion = ?
(a) Protium (b) Deuterium (c) Tritium (d) both b and c
Answer: (d)
Case Study 8: Isobars
Students were taught that elements with same mass number but different atomic numbers are called isobars. Example: ⁴⁰Ar (Z=18), ⁴⁰Ca (Z=20). They have different chemical properties but same mass.
Questions:
- Isobars have same:
(a) atomic number (b) mass number (c) neutrons (d) electrons
Answer: (b) - Atomic number of Ca = ?
(a) 18 (b) 19 (c) 20 (d) 21
Answer: (c) - Ar and Ca differ in:
(a) mass number (b) atomic number (c) both (d) none
Answer: (b) - Isobars have:
(a) identical chemical properties (b) different chemical properties (c) no properties (d) none
Answer: (b) - Isobars example = ?
(a) ¹²C and ¹³C (b) ⁴⁰Ar and ⁴⁰Ca (c) H and D (d) O and O₂
Answer: (b)
Case Study 9: Electronic Configuration
Riya arranged electrons of chlorine (Z=17) in shells: K=2, L=8, M=7. She understood that maximum electrons in a shell = 2n², where n = shell number. This helped her write configurations of other elements.
Questions:
- Electronic configuration of Cl = ?
(a) 2,8,7 (b) 2,7,8 (c) 8,7,2 (d) 2,8,8
Answer: (a) - Maximum electrons in M shell (n=3) = ?
(a) 8 (b) 18 (c) 32 (d) 2
Answer: (b) - Valency of Cl = ?
(a) 1 (b) 7 (c) 8 (d) 17
Answer: (a) - Which formula decides shell capacity?
(a) n² (b) 2n² (c) 2n (d) none
Answer: (b) - Chlorine belongs to group:
(a) 17 (b) 16 (c) 18 (d) 1
Answer: (a)
Case Study 10: Valency and Chemical Properties
In an activity, students compared sodium (Z=11) and magnesium (Z=12). Sodium configuration = 2,8,1 (valency 1), magnesium = 2,8,2 (valency 2). Both elements readily form positive ions (Na⁺, Mg²⁺) to complete octet, showing how valency governs chemical properties.
Questions:
- Valency of Na = ?
(a) 1 (b) 2 (c) 8 (d) 11
Answer: (a) - Valency of Mg = ?
(a) 1 (b) 2 (c) 12 (d) 8
Answer: (b) - Ion formed by sodium = ?
(a) Na⁻ (b) Na⁺ (c) Mg²⁺ (d) none
Answer: (b) - Both elements try to complete:
(a) duplet (b) triplet (c) octet (d) none
Answer: (c) - Valency determines:
(a) physical properties (b) chemical properties (c) isotopes (d) isobars
Answer: (b)