The ionisation energy of L i 2 + is 1.96 × 1 0 − 17 J ato m − 1 . The energy of the second excited state of H e + is :-

E 1 is proportional to Z 2 . E 1 ( L i 2 + ) = − 1.96 × 1 0 − 17 J. E 1 ( H e + ) = − 1.96 × 1 0 − 17 × 4/9 = − 8.71 × 1 0 − 18 J. The second excited state is n = 3 : E 3 = − 8.71 × 1 0 − 18 /9 = − 9.68 × 1 0 − 19 J. Traps: − 8.71 × 1 0 − 18 is the ground state; − 2.18 × 1 0 − 18 is the n = 2 level. Answer: (A).

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Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureBohr's Atomic ModelHard2 minQB

ChemistryHardsingle choice

The ionisation energy of $L i^{2 +}$ is $1.96 \times 1 0^{- 17} J ato m^{- 1}$ . The energy of the second excited state of $H e^{+}$ is :-

Options:

Answer:

Solution:

$E_{1}$ is proportional to $Z^{2} . E_{1} (L i^{2 +}) = - 1.96 \times 1 0^{- 17}$ J.

$E_{1} (H e^{+}) = - 1.96 \times 1 0^{- 17} \times 4/9 = - 8.71 \times 1 0^{- 18}$ J.

The second excited state is $n = 3 : E_{3} = - 8.71 \times 1 0^{- 18} /9 = - 9.68 \times 1 0^{- 19}$ J.

Traps: $- 8.71 \times 1 0^{- 18}$ is the ground state; $- 2.18 \times 1 0^{- 18}$ is the $n = 2$ level.

Answer: (A).

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Bohr's Atomic Model

⏱ 2mℹ️ Source: QB

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