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Chemistry - Atomic Structure Question with Solution | TestHub

ChemistryAtomic StructureSpectrumHard2 minQB

ChemistryHardnumerical

An electron of $a B e^{3 +}$ ion falls from the 3rd orbit to the 2nd orbit. The wavelength of the emitted photon, in nm (up to two decimal places), is $(E_{1}$ of $H = - 13.6$ eV, hc = 1240 eV nm)

Answer:

41.03

Solution:

$E = 13.6 \times Z^{2} (1/ n_{1}^{2} - 1/ n_{2}^{2}) = 13.6 \times 16 \times (1/4 - 1/9) = 217.6 \times 5/36 = 30.22$ eV.

$λ = 1240/30.22 = 41.03$ nm.

Answer: 41.03.

Stream:JEESubject:ChemistryTopic:Atomic StructureSubtopic:Spectrum

⏱ 2mℹ️ Source: QB

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