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← JEE Advanced 2018 · Paper 2
PhysicsNumerical+3Question 14 of 54

JEE Advanced 2018 · Paper 2, Physics question 14

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=2n = 2 to n=1n = 1 transition has energy 74.8 eV higher than the photon emitted in the n=3n = 3 to n=2n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV. The value of ZZ is __________.

Answer and worked solution

Official answer: 3

Photon energies for a hydrogen-like ion:
E2→1=13.6Z2⋅34,E3→2=13.6Z2⋅536.E_{2\to1}=13.6Z^2\cdot\frac34,\qquad E_{3\to2}=13.6Z^2\cdot\frac5{36}.

Difference.
13.6Z2(2736−536)=13.6Z2⋅2236=74.8  ⟹  Z2=74.8×3613.6×22=9.13.6Z^2\left(\frac{27}{36}-\frac5{36}\right)=13.6Z^2\cdot\frac{22}{36}=74.8\implies Z^2=\frac{74.8\times36}{13.6\times22}=9.

So Z=3Z=3.

Answer: 3