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The ionization energy of hydrogen atom is 13.6 eV. Following Bohr’s theory, the energy corresponding to a transition between 3rd and 4th orbit is

1. 3.40 eV
2. 1.51 eV
3. 0.85 eV
4. 0.66 eV

Correct answer: 0.66 eV

Solution

The energy difference between two orbits in a hydrogen atom is given by ΔE = 13.6 * (1/n1² - 1/n2²) eV. For n1 = 3 and n2 = 4, ΔE = 13.6 * (1/3² - 1/4²) = 13.6 * (1/9 - 1/16) = 13.6 * (7/144) ≈ 0.66 eV.

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