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

<p>To determine which element has the lowest first ionization enthalpy among the given options, we look at the atomic numbers:</p> <p><p>Atomic number 32 corresponds to Germanium (Ge).</p></p> <p><p>Atomic number 35 corresponds to Bromine (Br).</p></p> <p><p>Atomic number 87 corresponds to Francium (Fr).</p></p> <p><p>Atomic number 19 corresponds to Potassium (K).</p></p> <p>The element with the lowest first ionization energy is Francium (Fr) with atomic number 87. This can be represented by its electron configuration: $\text{Fr} - [\text{Rn}] 7s^1$. Francium, being a part of the alkali metal group, has the most loosely bound outer electron compared to the other elements listed, resulting in a lower first ionization enthalpy.</p>