The ionization enthalpy of hydrogen atom is 1.312 × 1 0 6 Jmol − 1 . The energy required to excite the electron in the atom from n = 1 to n = 2 is

Δ E = E 2 − E 1 = − 2 2 1.312 × 1 0 6 − ( − 1 1.312 × 1 0 6 ) = 9.84 × 1 0 5 J mol − 1

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ChemistryAtomic StructureSpectrumMedium2 minPYQ_2008

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The ionization enthalpy of hydrogen atom is $1.312 \times 1 0^{6} Jmol^{- 1}$ . The energy required to excite the electron in the atom from $n = 1$ to $n = 2$ is

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Answer:

Solution:

$Δ E = E_{2} - E_{1} = - \frac{1.312 \times 1 0 ^{6}}{2 ^{2}} - (- \frac{1.312 \times 1 0 ^{6}}{1}) = 9.84 \times 1 0^{5} J mol^{- 1}$

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Spectrum

⏱ 2mℹ️ Source: PYQ_2008

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