A hydrogen-like atom in higher state n emits photons of 10.2 eV and 17.0 eV in successive transitions reaching its first excited state. Alternatively, it emits 4.25 eV and 5.95 eV reaching second excited state. If Eₙ = −13.6Z²/n² eV, find n.
Physics · Mixed Chapter
JEE Main Practice

Question

A hydrogen-like atom in higher state n emits photons of 10.2 eV and 17.0 eV in successive transitions reaching its first excited state. Alternatively, it emits 4.25 eV and 5.95 eV reaching second excited state. If Eₙ = −13.6Z²/n² eV, find n.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Initial to n = 2 transition releases 10.2 + 17.0 = 27.2 eV. Initial to n = 3 transition releases 4.25 + 5.95 = 10.2 eV. Step 2 — Difference between final level energies is 17 eV = 13.6Z²(1/4 − 1/9) = (17/9)Z² eV. Thus Z² = 9. Step 3 — Eₙ = −13.6Z²/n². From ΔE(n→2) = 27.2: 27.2 = 122.4(1/4 − 1/n²). Step 4 — 1/n² = 1/4 − 2/9 = 1/36, so n = 6.