Explanation:

Millikan determined the elementary electric charge using the oil-drop experiment. Rutherford identified the atomic nucleus through alpha-particle scattering experiments.

Explanation:

Ionic atoms are formed by gaining or losing electrons, not protons.

Explanation:

Isoelectronic species have the same number of electrons. They do not necessarily have the same number of protons or neutrons.

Explanation:

Isotopes are atoms of the same element with the same atomic number but different neutron numbers. Isoelectronic species have the same number of electrons.

Pair I: ¹⁹₉F and ²¹₉F are isotopes.

Pair II: Ca²⁺ has 20 − 2 = 18 electrons; K⁺ has 19 − 1 = 18 electrons. They are isoelectronic.

Explanation:

Isotopes of the same element have the same numbers of protons and the same atomic number, but different numbers of neutrons and different mass numbers. Statements A–D are all correct, so E is the complete answer.

Explanation:

Both neutral strontium atoms have atomic number 38: each has 38 protons and 38 electrons. Their mass numbers are 88 and 90, so their neutron numbers are 50 and 52. A and B are both shared features; E includes both.

Explanation:

Protons	4 = (atomic number)	4 = (atomic number)
Electrons	4 = (atomic or proton number)	2 = (atomic number – cations)    = 4 – 2
Neutrons	5 = (mass number – atomic number)	5 = (mass number – atomic number)
Mass number	9	9

Solution:

Neutron numbers

= mass number – atomic or proton numbers

= 79 – 34

= 45

Solution:

In an ionic atom: number of protons = atomic number ≠  number of electrons

30 protons → atomic number is 30 → ₃₀Zn

mass number = proton numbers + neutron numbers

= 30 + 35

= 65

= ⁶⁵₃₀Zn

30 protons and 28 electrons indicates the loss of 2 electrons → ⁶⁵₃₀Zn²⁺

Solution:

Magnesium ion, Mg²⁺: 12 – 2 = 10

Fluoride, F^–: 9 + 1 = 10

Oxide, O^2-: 8 + 2 = 10

Neon, Ne : 10

Fluorine, F : 9

Solution:

Se^2- : 34 atomic number (protons) + 2 electrons

= 36 electrons

Rb⁺ : 37 atomic number (protons) – 1 electron

= 36 electrons

Solution:

₁₆S^2- : 16 atomic number (protons) + 2 electrons

= 18 electrons

₁₇Cl^– : 17 atomic number (protons) + 1 electron

= 18 electrons

₁₈Ar : 18 atomic number (protons)

= 18 electrons

₂₁Sc⁺ : 21 atomic number (protons) – 1 electron

= 20 electrons

₂₀Ca²⁺ : 20 atomic number ( protons) – 2 electrons

= 18 electrons

Solution:

The balanced nuclear equation is ²⁰⁰₈₀Hg → ⁴₂He + ¹⁹⁶₇₈Pt.

Atomic number is conserved: 80 = 2 + 78.

Mass number is conserved: 200 = 4 + 196.

The daughter nucleus is platinum-196, with 196 − 78 = 118 neutrons. Alpha emission decreases the mass number by 4 and the atomic number by 2. Therefore, D is the only incorrect statement.

Solution:

¹⁹⁰₇₆Os  →  ⁴₂α + ⁰₋₁β⁻ + ⁰₀γ + ^b_aX

The sum of the reactant atomic numbers must be equal to the sum of the product atomic numbers.

76 = 2 + (-1) + 0 + a, a = 75

Proton number of X = 75 → Re

The sum of the reactant mass numbers must be equal to the sum of the product mass numbers.

190 = 4 + 0 + 0 + b, b = 186

Mass number of X = 186

Therefore, the daughter element X is ¹⁸⁶₇₅Re.

Solution:

¹⁸⁸₇₄W → ⁴₂He + 2(⁰₋₁β⁻) + 3(⁰₀γ) + ¹⁸⁴₇₄W

Parent atomic number = 2 + 2(−1) + 3(0) + 74 = 74.

Parent mass number = 4 + 2(0) + 3(0) + 184 = 188.

Therefore, the parent nucleus is tungsten-188, ¹⁸⁸₇₄W.

Solution:

₇₆Os → x(⁴₂α) + 2(⁰₋₁β⁻ )+ ₇₂Hf

76 = x(2) + 2(-1) +72

x = 3

Explanation:

Alpha (α) decay decreases the mass number by 4 and the atomic number by 2.

Beta-minus (β⁻) decay increases the atomic number by 1 without changing the mass number.

Gamma (γ) emission changes neither the mass number nor the atomic number. Therefore, I: α, II: β⁻, III: γ (A).