The atomic number of the element from the following with lowest 1st ionisation enthalpy is:

🧠 Lowest IE = most metallic = bottom-left of the periodic table. Among the given Z values, that is Fr ($Z = 87$) Ionisation enthalpy decreases down a group (more shells, weaker pull) and decreases right-to-left across a period. The very bottom-left of the table is occupied by Group 1 alkali metals — and the heaviest of these (Fr) has the lowest $IE_1$ of all elements.

🗺️ Identify each element $Z = 32$: Germanium (Ge) — $p$-block metalloid, Group 14, Period 4. $IE_1 \approx 762\,kJ/mol$. $Z = 35$: Bromine (Br) — halogen, Group 17. High IE. $IE_1 \approx 1140\,kJ/mol$. $Z = 19$: Potassium (K) — Group 1, Period 4. $IE_1 \approx 419\,kJ/mol$. $Z = 87$: Francium (Fr) — Group 1, Period 7. $IE_1 \approx 380\,kJ/mol$ (lowest of all naturally occurring elements).

Among these, Fr has the lowest $IE_1$.

⚠️ The trap Many students pick K because it is a known alkali metal with a low IE. K is close (419), but Fr is even lower (380). When the choice is between two alkali metals, the heavier one always wins for low IE.

$\boxed{\text{Answer: (c)}}$