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JEE Main 2019 (12 Apr, Shift 1) · Transitions, Photon Emission & Excitation

Transitions, Photon Emission & Excitation question from JEE Main 2019 (12 Apr, Shift 1)

Q57JEE Main 2019 · 12 APR · SHIFT 1Difficult

Transitions, Photon Emission & Excitation

An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength λ\lambda, energy E=1240 eVλ(in nm)E = \frac{1240\text{ eV}}{\lambda(\text{in nm})}):

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Answer: (c)

  1. E1=1240108.5=11.43E_1 = \frac{1240}{108.5} = 11.43 eV, E2=124030.4=40.79E_2 = \frac{1240}{30.4} = 40.79 eV; total =52.22= 52.22 eV.
  2. For He+^+, En=−54.4n2E_n = -\frac{54.4}{n^2} eV, ground state −54.4-54.4 eV.
  3. En=−54.4+52.22=−2.18E_n = -54.4 + 52.22 = -2.18 eV ⇒n2=54.42.18≈25\Rightarrow n^2 = \frac{54.4}{2.18} \approx 25
  4. So n=5n = 5, the answer is (c).
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