Explanation:

Millikan measured the charge of the electron in the oil-drop experiment. Rutherford discovered the nucleus through alpha-particle scattering.

Explanation:

Ions are formed by gaining or losing electrons, not protons.

Explanation:

Isoelectronic species have the same number of electrons. Their neutron numbers may be the same or different; different elements have different proton numbers.

Explanation:

Pair I consists of isotopes: atoms of the same element with the same number of protons but different numbers of neutrons and different mass numbers. Pair II is isoelectronic: both ions have the same number of electrons.

Ca²⁺: 20 electrons –2 electrons = 18 electrons

K⁺: 19 electrons – 1 electron = 18 electrons

Explanation:

Isotopes of the same element have the same number of protons and the same atomic number, but different numbers of neutrons and different mass numbers. All statements A–D are correct.

Explanation:

Neutral atoms have the same numbers of protons and electrons. Both strontium atoms have 38 protons and 38 electrons. Their mass numbers are 88 and 90, and their neutron numbers are 50 and 52, respectively. Thus, A and B are the shared features.

Explanation:

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

Solution:

Number of neutrons

= mass number – atomic number

= 79 – 34

= 45

Solution:

For a monatomic ion: number of protons = atomic number ≠  number of electrons

30 protons → atomic number is 30 → ₃₀Zn

mass number = number of protons + number of neutrons

= 30 + 35

= 65

= ⁶⁵₃₀Zn

30 protons and 28 electrons indicate 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:

A. ²⁰⁰₈₀Hg undergoes alpha emission in the reaction shown.

B. The sum of the reactant atomic numbers (80) must be equal to the sum of the product atomic numbers (2+a).
²⁰⁰₈₀Hg → ⁴₂He + ^b_aX

80 = 2 + a, a = 78

Atomic number of X = 78 → Pt (platinum)

C. The sum of the reactant mass numbers must equal the sum of the product mass numbers.

200 = 4 + b, b = 196

Mass number of X = 196

Neutron number of X: 196 – 78 = 118

D. The mass number of X is decreased by 4.

E. Atomic, neutron, and mass numbers for X are changed.

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:

^b_aX → 1(⁴₂α) + 2(⁰_-1β) + 3(⁰₀γ) + ¹⁸⁴₇₄W

a = 1(2)+2(-1)+3(0)+74

a = 74, tungsten

b = 1(4) + 2(0) + 3(0) +184

b = 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:  parent  → ⁴₂He + daughter

Beta-minus (β⁻) decay:  parent → ⁰_-1e + daughter

Gamma (γ) decay:  parent  → ⁰₀γ  + daughter