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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Physics
Chapter
Atomic Physics

JEE Advanced 2018 · Paper 2 · Physics · Atomic Physics

Consider a hydrogen-like ionized atom with atomic number Z with a...

Consider a hydrogen-like ionized atom with atomic number ZZ with a single electron. In the emission spectrum of this atom, the photon emitted in the n=2ton=1n = 2ton = 1 transition has energy 74.8eV74.8eV higher than the photon emitted in the n=3ton=2n = 3ton = 2 transition. Given that the ionization energy of the hydrogen atom is 13.6eV13.6eV, what is the value of ZZ?

Correct answer

0

Step-by-step solution

∆E2→1=13.6×Z2[1−14]=13.6×Z2[34]∆ {E}_{2 \to 1} = 13.6 \times {Z}^{2}\left[1 - \dfrac{1}{4}\right] = 13.6 \times {Z}^{2}\left[\dfrac{3}{4}\right]
∆E3→2=13.6×Z2[14−19]=13.6×Z2[536]∆ {E}_{3 \to 2} = 13.6 \times {Z}^{2}\left[\dfrac{1}{4} - \dfrac{1}{9}\right] = 13.6 \times {Z}^{2}\left[\dfrac{5}{36}\right]
∆E2→1=∆E3−2+74.8∆ {E}_{2 \to 1} = ∆ {E}_{3 - 2} + 74.8
13.6×Z2[34]=13.6×Z2[536]+74.813.6 \times {Z}^{2}\left[\dfrac{3}{4}\right] = 13.6 \times {Z}^{2}\left[\dfrac{5}{36}\right] + 74.8
13.6×Z2[34−536]=74.813.6 \times {Z}^{2}\left[\dfrac{3}{4} - \dfrac{5}{36}\right] = 74.8
Z2=9{Z}^{2} = 9
Z=+3Z = + 3
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