Question:
Hydrogen (H), deuterium (D), singly ionized helium (He⁺) and doubly ionized lithium (Li²+) all have one electron around the nucleus. Consider n=2 to n=1 transition. The wavelengths of emitted radiations are lambda₁, lambda₂, lambda₃ and lambda₄ respectively. Then approximately
Hydrogen (H), deuterium (D), singly ionized helium (He⁺) and doubly ionized lithium (Li²+) all have one electron around the nucleus. Consider n=2 to n=1 transition. The wavelengths of emitted radiations are lambda₁, lambda₂, lambda₃ and lambda₄ respectively. Then approximately
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For hydrogen-like species, wavelength varies inversely as Z².
Updated On: Mar 20, 2026
- \(\lambda_1=\lambda_2=4\lambda_3=9\lambda_4\)
- \(4\lambda_1=2\lambda_2=2\lambda_3=\lambda_4\)
- \(\lambda_1=2\lambda_2=2\lambda_3=3/2\,\lambda_4\)
- lambda₁=lambda₂=2lambda₃=3√(2)lambda₄
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The Correct Option is A
Solution and Explanation
Step 1: For hydrogen-like atoms λ ∝ (1)/(Z²)
Step 2: Atomic numbers H, D: Z=1; He⁺: Z=2; Li²+: Z=3
Step 3: Hence lambda₁=lambda₂=(1)/(1²), lambda₃=(1)/(4), lambda₄=(1)/(9)
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