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NEET PHYSICSMedium

Hydrogen (1H^1H), Deuterium (2H^2H), singly ionised helium (He+He^+), and doubly ionised lithium (Li2+Li^{2+}) all have one electron around the nucleus. Consider an electron transition from n=2 to n=1. If the wavelengths of emitted radiations are λ1\lambda_1, λ2\lambda_2, λ3\lambda_3, and λ4\lambda_4 respectively, then approximately which one of the following is correct?

A

4\lambda ₁ = 2\lambda ₂ = 2\lambda ₃ = \lambda ₄

B

\lambda ₁ = 2\lambda ₂ = 2\lambda ₃ = \lambda ₄

C

\lambda ₁ = \lambda ₂ = 4\lambda ₃ = 9\lambda ₄

D

\lambda ₁ = 2\lambda ₂ = 3\lambda ₃ = \lambda ₄

Step-by-Step Solution

According to the Bohr model for hydrogen-like species (one-electron systems), the energy of a transition is given by ΔEZ2(1n121n22)\Delta E \propto Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right). Since the wavelength λ\lambda is inversely proportional to energy (ΔE=hc/λ\Delta E = hc/\lambda), the relationship is λ1Z2\lambda \propto \frac{1}{Z^2} for the same transition (n=21n=2 \to 1).

  1. Hydrogen (1H^1H): Z=1Z=1, so λ11\lambda_1 \propto 1.
  2. Deuterium (2H^2H): Z=1Z=1 (isotope effects on Rydberg constant are negligible for this approximation), so λ2λ1\lambda_2 \approx \lambda_1.
  3. Helium ion (He+He^+): Z=2Z=2, so λ3122=14\lambda_3 \propto \frac{1}{2^2} = \frac{1}{4}. This implies λ3=λ14\lambda_3 = \frac{\lambda_1}{4} or λ1=4λ3\lambda_1 = 4\lambda_3.
  4. Lithium ion (Li2+Li^{2+}): Z=3Z=3, so λ4132=19\lambda_4 \propto \frac{1}{3^2} = \frac{1}{9}. This implies λ4=λ19\lambda_4 = \frac{\lambda_1}{9} or λ1=9λ4\lambda_1 = 9\lambda_4.

Combining these, we get λ1=λ2=4λ3=9λ4\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4.

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