JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

What is the energy (in J atom $^{-1}$ ) required for the following process?

$Li^{2+}(g) \rightarrow Li^{3+}(g) + e^-$

(Take the ionization energy for the H atom in the ground state as $2.18 \times 10^{-18}$ J atom $^{-1}$ )

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$8.72 \times 10^{-18}$
B
$1.962 \times 10^{-18}$
C
$1.962 \times 10^{-17}$
D
$6.54 \times 10^{-17}$

Did you get this right?

Solution

The given process $Li^{2+}(g) \rightarrow Li^{3+}(g) + e^-$ represents the ionization of the hydrogen-like ion $Li^{2+}$ from its ground state.

The ionization energy for a hydrogen-like species is given by the formula:

$IE = IE_{H} \times Z^2$

For $Li^{2+}$ , the atomic number $Z = 3$ .

Given the ionization energy of the hydrogen atom $IE_{H} = 2.18 \times 10^{-18}$ J atom $^{-1}$ .

Substituting the values:

$IE = 2.18 \times 10^{-18} \times (3)^2$

$IE = 2.18 \times 10^{-18} \times 9$

$IE = 19.62 \times 10^{-18}$ J atom $^{-1}$

$IE = 1.962 \times 10^{-17}$ J atom $^{-1}$

Answer: $1.962 \times 10^{-17}$

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

What is the energy (in J atom $^{-1}$ ) required for the following process?

$Li^{2+}(g) \rightarrow Li^{3+}(g) + e^-$

(Take the ionization energy for the H atom in the ground state as $2.18 \times 10^{-18}$ J atom $^{-1}$ )

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$8.72 \times 10^{-18}$
B
$1.962 \times 10^{-18}$
C
$1.962 \times 10^{-17}$
D
$6.54 \times 10^{-17}$

Did you get this right?

Solution

The given process $Li^{2+}(g) \rightarrow Li^{3+}(g) + e^-$ represents the ionization of the hydrogen-like ion $Li^{2+}$ from its ground state.

The ionization energy for a hydrogen-like species is given by the formula:

$IE = IE_{H} \times Z^2$

For $Li^{2+}$ , the atomic number $Z = 3$ .

Given the ionization energy of the hydrogen atom $IE_{H} = 2.18 \times 10^{-18}$ J atom $^{-1}$ .

Substituting the values:

$IE = 2.18 \times 10^{-18} \times (3)^2$

$IE = 2.18 \times 10^{-18} \times 9$

$IE = 19.62 \times 10^{-18}$ J atom $^{-1}$

$IE = 1.962 \times 10^{-17}$ J atom $^{-1}$

Answer: $1.962 \times 10^{-17}$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

Browse all Chemistry topics

JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry