A double charged lithium atom is equivalent to hydrogen whose atomic number is 3. The wavelength of required radiation for emitting electron from first to third Bohr orbit in Li++ will be (Ionisation energy of hydrogen atom is 13.6 eV)
A
182.51∘A
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B
177.17∘A
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C
142.25∘A
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D
113.74∘A
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Solution
The correct option is D113.74∘A Energy of a electron in nth orbit of a hydrogen like atom is given by En=−13.6Z2n2eV, and Z = 3 for Li Required energy for said transition ΔE=E3−E1=13.6Z2(112−132)=13.6×32[89]=108.8eV=108.8×1.6×10−19J Now using ΔE=hcΔE⇒λ=6.6×10−34×3×108108.8×1.6×10−19=0.11374×10−7m⇒λ=113.74∘A