Question

isotopes
fill in the missing information about the different isotopes of different elements. if all the answers in a row are correct, then the row should change color.
columns: element symbol, element name, atomic number, # protons, # neutrons, # electrons, mass number, isotope notation, isotope name.
rows labeled 5, 6, 7, 8, 9, 10.
entries include iodine, lead-207, numbers 6, 11, 12, 20, 74, etc.
isotope notation section shows atomic #, mass #, element symbol.

Explanation:

To solve problems related to isotopes, we use the following relationships:

• Atomic Number (\(Z\)) = Number of Protons (\(p^+\)) = Number of Electrons (\(e^-\)) (in neutral atoms).
• Mass Number (\(A\)) = Number of Protons (\(p^+\)) + Number of Neutrons (\(n\)).

Example 1: Carbon (Atomic Number = 6)

Step 1: Find Protons and Electrons

Atomic Number (\(Z\)) = 6, so \(p^+ = 6\) and \(e^- = 6\) (neutral atom).

Step 2: Find Neutrons

Mass Number (\(A\)) = 12 (given).
Neutrons (\(n\)) = \(A - Z = 12 - 6 = 6\).

Example 2: Iodine (Atomic Number for Iodine is 53)

Step 1: Find Protons and Electrons

Atomic Number (\(Z\)) = 53, so \(p^+ = 53\) and \(e^- = 53\) (neutral atom).

Step 2: Find Neutrons

Mass Number (\(A\)) = 74 + 53 = 127 (since Neutrons = 74, \(A = 53 + 74 = 127\)).

Example 3: Lead-207 (Isotope \(^{207}\text{Pb}\))

Step 1: Find Protons and Electrons

Atomic Number for Lead (\(Z\)) = 82, so \(p^+ = 82\) and \(e^- = 82\) (neutral atom).

Step 2: Find Neutrons

Mass Number (\(A\)) = 207.
Neutrons (\(n\)) = \(A - Z = 207 - 82 = 125\).

Example 4: Element with \(p^+ = 9\), \(n = 10\)

Step 1: Find Atomic Number and Electrons

\(Z = p^+ = 9\), so \(e^- = 9\) (neutral atom).

Step 2: Find Mass Number

\(A = p^+ + n = 9 + 10 = 19\).

Step 3: Identify Element

Atomic Number 9 = Fluorine (F).

Filling the Table (Key Rows):

Element Symbol	Element Name	Atomic Number	# Protons	# Neutrons	# Electrons	Mass Number	Isotope Notation	Isotope Name
C	Carbon	6	6	6	6	12	\(^{12}\text{C}\)	Carbon-12
I	Iodine	53	53	74	53	127	\(^{127}\text{I}\)	Iodine-127
Pb	Lead	82	82	125	82	207	\(^{207}\text{Pb}\)	Lead-207

For any missing cell, apply \( Z = p^+ = e^- \) and \( A = p^+ + n \) to find the unknowns.

Answer:

To solve problems related to isotopes, we use the following relationships:

• Atomic Number (\(Z\)) = Number of Protons (\(p^+\)) = Number of Electrons (\(e^-\)) (in neutral atoms).
• Mass Number (\(A\)) = Number of Protons (\(p^+\)) + Number of Neutrons (\(n\)).

Example 1: Carbon (Atomic Number = 6)

Step 1: Find Protons and Electrons

Atomic Number (\(Z\)) = 6, so \(p^+ = 6\) and \(e^- = 6\) (neutral atom).

Step 2: Find Neutrons

Mass Number (\(A\)) = 12 (given).
Neutrons (\(n\)) = \(A - Z = 12 - 6 = 6\).

Example 2: Iodine (Atomic Number for Iodine is 53)

Step 1: Find Protons and Electrons

Atomic Number (\(Z\)) = 53, so \(p^+ = 53\) and \(e^- = 53\) (neutral atom).

Step 2: Find Neutrons

Mass Number (\(A\)) = 74 + 53 = 127 (since Neutrons = 74, \(A = 53 + 74 = 127\)).

Example 3: Lead-207 (Isotope \(^{207}\text{Pb}\))

Step 1: Find Protons and Electrons

Atomic Number for Lead (\(Z\)) = 82, so \(p^+ = 82\) and \(e^- = 82\) (neutral atom).

Step 2: Find Neutrons

Mass Number (\(A\)) = 207.
Neutrons (\(n\)) = \(A - Z = 207 - 82 = 125\).

Example 4: Element with \(p^+ = 9\), \(n = 10\)

Step 1: Find Atomic Number and Electrons

\(Z = p^+ = 9\), so \(e^- = 9\) (neutral atom).

Step 2: Find Mass Number

\(A = p^+ + n = 9 + 10 = 19\).

Step 3: Identify Element

Atomic Number 9 = Fluorine (F).

Filling the Table (Key Rows):

Element Symbol	Element Name	Atomic Number	# Protons	# Neutrons	# Electrons	Mass Number	Isotope Notation	Isotope Name
C	Carbon	6	6	6	6	12	\(^{12}\text{C}\)	Carbon-12
I	Iodine	53	53	74	53	127	\(^{127}\text{I}\)	Iodine-127
Pb	Lead	82	82	125	82	207	\(^{207}\text{Pb}\)	Lead-207

For any missing cell, apply \( Z = p^+ = e^- \) and \( A = p^+ + n \) to find the unknowns.