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Chemistry Question 52 – JEE-MAIN 2026

What is the energy (in J atom1) required for the following process ? Li2+(g) Li3+(g) + e (Take the ionization energy for the H atom in the ground state as 2.18×1018 J atom1)

The process involves the ionization of a Li2+ ion, which is a hydrogen-like species (contains only one electron).

Step 1: Identify the Species and Process✦ Active

The given process is Li2+(g) Li3+(g) + e. This represents the removal of the last electron from a Li2+ ion. Lithium (Li) has an atomic number Z=3. A Li2+ ion has lost two electrons, leaving it with only one electron, making it a hydrogen-like species.

Step 2: Recall the Formula for Ionization Energy of Hydrogen-like Species○ Expand

For a hydrogen-like species, the ionization energy from the ground state (n=1) is directly proportional to the square of its atomic number (Z). The formula is given by:

IE=IEH×Z2

where IEH is the ionization energy of the hydrogen atom in its ground state.

Step 3: Calculate the Energy Required○ Expand

Given IEH=2.18×1018 J atom1 and for Lithium, Z=3. Substitute these values into the formula:

IELi2+=(2.18×1018 J atom1)×(3)2
IELi2+=(2.18×1018)×9 J atom1
IELi2+=19.62×1018 J atom1

Converting to a standard scientific notation format:

IELi2+=1.962×1017 J atom1

This value matches option 3.

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