Explanation:

Millikan measured the mass of electron by oil-drop experiment. Discovery of nucleus by the scattering alpha (α) particles is Rutherford.

Explanation:

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

Explanation:

Isoelectronics have the same number of electrons but different number of protons.

Explanation:

An isotope has the same number of atoms (protons) but different numbers of mass (neutrons). An isoelectronic pair has the same number of electrons between two different atoms.

Ca²⁺: 20 electrons –2 electrons = 18 electrons

K⁺: 19 electrons – 1 electrons = 18 electrons

Explanation:

Isotopes have the same number of protons, but these have the different number of mass (neutrons).

Explanation:

Neutral atoms have the same numbers of protons and electrons. These two atoms have different atomic mass (88 and 90) and different numbers of neutrons.

Explanation:

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

Solution:

Neutron numbers

= atomic mass – 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

atomic mass = 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

₁₆Cl^– : 17 atomic number (protons) + 1 electron

= 18 electrons

₁₆Ar : 18 atomic number (protons)

= 18 electrons

₁₆Sc⁺ : 21 atomic number (protons) – 1 electron

= 20 electrons

₁₆Ca²⁺ : 20 atomic number ( protons) – 2 electrons

= 18 electrons

Solution:

A. ²⁰⁰₈₀Hg is alpha emission.

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

80 = 2 + a, a = 78

Atomic number of X = 78 à Pt (platinum)

C. The sum of the reactant atomic numbers must be equal to the sum of the product atomic 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 ¹⁸⁶₇₇Re.

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 known nucleus is Tungsten-188, ¹⁸⁸₇₄X .

Solution:

₇₆Os → x(⁴₂α) + 2(⁰_-1β )+ ₇₂Hf

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

x = 3

Explanation:

Alpha (α) decay:  parent  → ⁴₂He + daughter

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

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