The energy of second orbit of hydrogen atom is − 5.45 × 1 0 − 19 J . What is the energy of first orbit of Li 2 + ion (in J ) ?

E n = − 13.6 [ n 2 z 2 ] e V ⇒ E 1 = − 13.6 [ 1 2 3 2 ] eV = − 122.4 eV = − 122.4 × 1.602 × 1 0 − 19 J = − 1.962 × 1 0 − 17 J

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ChemistryAtomic StructureBohr's Atomic ModelMedium2 minPYQ_2024

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The energy of second orbit of hydrogen atom is $- 5.45 \times 1 0^{- 19} J$ . What is the energy of first orbit of $Li^{2 +}$ ion (in J $)$ ?

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Answer:

Solution:

$E_{n} = - 13.6 [\frac{z ^{2}}{n ^{2}}] e V$ $\Rightarrow E_{1} = - 13.6 [\frac{3 ^{2}}{1 ^{2}}] eV = - 122.4 eV$ $= - 122.4 \times 1.602 \times 1 0^{- 19} J = - 1.962 \times 1 0^{- 17} J$

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

⏱ 2mℹ️ Source: PYQ_2024

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