Modern periodic law — based on atomic number: Worked examples

Before anything, let us re-earn the one tool we lean on, because we will use it in almost every example.

The scenario matrix

Every question in this topic falls into one of these cells. The worked examples below are labelled with the cell they cover, so together they leave no gap.

Cell	Case class	What makes it tricky	Example
A	Straight line vs mass	Choose the correct axis	Ex 1
B	Anomalous pair (mass ↔ chemistry conflict)	Order flips vs mass	Ex 2
C	Find $Z$ from a frequency ratio	Constant $a$ cancels	Ex 3
D	Find $a$ and $b$ from two data points	Solve 2 equations	Ex 4
E	Predict a missing/unknown element's frequency	Forward substitution	Ex 5
F	Degenerate input: $Z = b$ (line hits the axis)	$ν = 0$ , physical meaning	Ex 6
G	Limiting/sign check: what if you forget $b$ ?	Small- $Z$ error blows up	Ex 7
H	Real-world word problem	Translate chemistry → $Z$	Ex 8
I	Exam twist: wavelength given, not frequency	$ν = c / λ$ conversion	Ex 9

Worked Examples

Cell A — which axis gives the straight line?

Worked example Example 1

You measure $ν$ for calcium, titanium and iron and want a clean straight-line plot. Do you put atomic number or atomic mass on the horizontal axis?

Forecast: Guess now — mass or number?

Step 1. Write the law: $ν = a (Z - b)$ . Why this step? The variable that appears inside the equation is the one that plots straight.

Step 2. The independent variable is $Z$ (atomic number), and it grows in perfect steps of $1$ . Mass grows irregularly because neutron counts jump around (isotopes, packing). So mass would give a wobbly, non-linear scatter. Why this step? A straight line needs a variable that changes evenly; only $Z$ does.

Answer: Plot against atomic number.

Verify: With $Z$ on the axis the points fall dead-straight (see figure). Ar–K, the classic mass anomaly, would even go backwards on a mass axis — proof mass is not the true order.

Figure — Modern periodic law — based on atomic number

Cell B — the anomalous pair

Cell C — find $Z$ from a frequency ratio

Cell D — find both constants $a$ and $b$

Cell E — predict an unknown element

Cell F — the degenerate input $Z = b$

Worked example Example 6

On a line where $b = 1$ , what does the law predict for a hypothetical element with $Z = 1$ (hydrogen)? What does the graph do there?

Forecast: Plug $Z = b$ — what happens to $ν$ ?

Step 1. $ν = a (Z - b) = a (1 - 1) = 0$ , so $ν = 0$ . Why this step? When $Z = b$ the bracket vanishes — this is the point where the straight line crosses the $Z$ -axis.

Step 2. Physical reading: $ν = 0$ means no characteristic X-ray of that series. Hydrogen has only one electron, so there is no inner-to-inner K-shell transition to emit the line at all. Why this step? The maths hitting zero warns us the physical process disappears — a degenerate case, not a real spectral line.

Answer: $ν = 0$ ; the line meets the axis at $Z = b$ , and no such X-ray exists.

Verify: This is exactly why the plot does not pass through the origin ( $Z = 0$ ) but through $Z = b$ . Confirms the " $ν = a (Z - b)$ , keep the $b$ " rule — see Effective Nuclear Charge and Shielding.

Cell G — limiting/sign trap: dropping $b$

Cell H — real-world word problem

Cell I — exam twist: wavelength instead of frequency

Worked example Example 9

An X-ray line has wavelength $λ = 1.5 \times 1 0^{- 10}$ m (given, not frequency!). On a line with $a = 5.0 \times 1 0^{7}$ , $b = 1$ , find $Z$ . Use speed of light $c = 3.0 \times 1 0^{8}$ m/s.

Forecast: You cannot use $λ$ directly — what conversion do you need first?

Step 1. Convert wavelength to frequency: $ν = \frac{c}{λ}$ . Why this step? Moseley's law is written in $ν$ ; a wave's frequency and wavelength are tied by $ν λ = c$ (they multiply to the wave's speed). $ν = \frac{3.0 \times 1 0 ^{8}}{1.5 \times 1 0 ^{- 10}} = 2.0 \times 1 0^{18} Hz .$

Step 2. Take the square root: $ν = 2.0 \times 1 0^{18} \approx 1.414 \times 1 0^{9}$ . Why this step? The law needs $ν$ , not $ν$ .

Step 3. Solve for $Z$ : $Z = \frac{1.414 \times 1 0 ^{9}}{5.0 \times 1 0 ^{7}} + 1 \approx 28.28 + 1 = 29.28$ . Why this step? Same rearrangement as Ex 8.

Step 4. $Z$ must be a whole number → round to $Z = 29$ (copper). Why this step? Atomic number is an integer; the tiny decimal is rounding/measurement slack.

Answer: $Z \approx 29$ — copper.

Verify: Copper's K-line wavelength really is ~ $1.54 \times 1 0^{- 10}$ m, gratifyingly close to our input. ✓

Recall Quick self-test

A sample gives $ν = 1.2 \times 1 0^{9}$ on the line $a = 5 \times 1 0^{7}, b = 1$ . Element? ::: $Z = 25$ , manganese. Why doesn't the $ν$ -vs- $Z$ line pass through the origin? ::: It crosses at $Z = b$ , not $Z = 0$ , because of shielding. Given $λ$ , what must you do before using Moseley's law? ::: Convert to frequency with $ν = c / λ$ .

Connections

1.4.02 Modern periodic law — based on atomic number (Hinglish) — the parent topic
Moseley's X-ray Experiment — where the law comes from
Effective Nuclear Charge and Shielding — meaning of the constant $b$
Atomic Number and Atomic Mass — why $Z$ orders elements
Bohr Model of the Atom — source of the $(Z - b)^{2}$ energy law
Mendeleev's Periodic Table — the mass-based order these anomalies broke
Periodic Trends — the repeating properties themselves