KVPY SB SX Exam 2011 Question Paper

Question : 104 of 120

Marks: +1, -0

The electron in hydrogen atom is in the First Bohr orbit (n = 1). The ratio o f transition energies. E(n = 1→ n = 3) to E(n = 1 → n = 2) is

Solution:

\ΔE=13.6(1/n_1^2-1/n_2^2)\eV\/\atom

{\ΔE_{1→3}}/{\ΔE_{1→2}}={1/1^2-1/3^2}/{1/1^2-1/2^2}=32/27

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