A hydrogen-like He + ion undergoes a transition from n = 2 to n = 1. Using R = 1.097 10 7 m -1 , what is the approximate wavelength of the emitted radiation?

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A hydrogen-like He+ ion undergoes a transition from n = 2 to n = 1. Using R = 1.097 × 107 m-1, what is the approximate wavelength of the emitted radiation?

Answer: 30.4 nm.

A 121.6 nm
B 91.2 nm
C 30.4 nm
D 243.0 nm

Correct answer: C. 30.4 nm

Explanation: For He+, Z = 2. Using 1/λ = RZ2(1/12 - 1/22), the wavelength is approximately 30.4 nm.

Comparison of the Thomson model and the Rutherford model of the atom: in the Thomson plum pudding model alpha particles pass almost straight through a diffuse positive sphere, while in the Rutherford model a concentrated central nucleus deflects some alpha particles through large angles, with the lower panels showing the alpha particle source and gold foil and the observed result of wide-angle scattering

The Geiger–Marsden alpha-scattering result: a few alpha particles bounce back at large angles, which rules out Thomson’s diffuse model and demands a tiny, massive, positively charged nucleus. Image: Kurzon, CC BY 3.0, via Wikimedia Commons.

Concept context

Thomson's plum-pudding model, Rutherford's nuclear model from alpha-scattering, Bohr's postulates and hydrogen spectrum, spectral series (Lyman to Pfund), Rydberg formula, and limitations of each model.

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A hydrogen-like He<sup>+</sup> ion undergoes a transition from n = 2 to n = 1. Using R = 1.097 &times; 10<sup>7</sup> m<sup>-1</sup>, what is the approximate wavelength of the emitted radiation?

Concept context

More Atoms questions

A hydrogen-like He<sup>+</sup> ion undergoes a transition from n = 2 to n = 1. Using R = 1.097 × 10<sup>7</sup> m<sup>-1</sup>, what is the approximate wavelength of the emitted radiation?