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In a hydrogen-like species, an electron transitions from an orbital having 2 radial nodes and 2 angular nodes to the orbital where the wave function has the same sign in all directions at every distance. If the energy of the emitted photon is 326.4 eV, identify the atom or ion.
- H
- Li²+
- He⁺
- B⁴+
Correct answer: B⁴+
Solution
Orbital with 2 angular nodes: l=2. Radial nodes = n-l-1 = 2 => n=5. The 1s orbital (n=1, l=0) has a wave function that is positive everywhere and spherically symmetric. Energy emitted = 13.6*Z²*(1/1 - 1/25) = 13.6*Z²*(24/25) = 326.4 eV. So Z² = 326.4*25/(13.6*24) = 8160/326.4 = 25 => Z=5 => Boron (B⁴+).
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