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

• A. \(\lambda_1=\lambda_2=4\lambda_3=9\lambda_4\)
• B. \(4\lambda_1=2\lambda_2=2\lambda_3=\lambda_4\)
• C. \(\lambda_1=2\lambda_2=2\lambda_3=3/2\,\lambda_4\)
• D. lambda₁=lambda₂=2lambda₃=3√(2)lambda₄

Hint:

For hydrogen-like species, wavelength varies inversely as Z².

Answer 1

The Correct Option is A

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)