Hydrogen atom: Doubly-ionized lithium is a hydrogen-like ion, in which the nucleus with charge 3e is orbited by a single electron. What wavelength light would be emitted from Li⁺⁺ as the electron dropped from the n = 4 to the n = 3 orbit?

Question

Quizplus · Accepted Answer

The Rydberg formula for the wavelength of light emitted by a hydrogen-like ion is:$$\frac{1}{\lambda} = R_H * (\frac{1}{n_f^2} - \frac{1}{n_i^2})$$where:* $$\lambda$$ is the wavelength of light emitted* $$R_H$$ is the Rydberg constant for hydrogen ($$1.0973731	imes10^7 \ m^{-1}$$)* $$n_f$$ is the final energy level of the electron* $$n_i$$ is the initial energy level of the electronFor doubly-ionized lithium, $$Z = 3$$, so the Rydberg constant is:$$R_{Li^{++}} = \frac{R_H}{Z^2} = \frac{1.0973731	imes10^7 \ m^{-1}}{3^2} = 1.226	imes10^6 \ m^{-1}$$Substituting the given values into the Rydberg formula, we get:$$\frac{1}{\lambda} = 1.226	imes10^6 \ m^{-1} * (\frac{1}{3^2} - \frac{1}{4^2}) = 4.80	imes10^4 \ m^{-1}$$Therefore, the wavelength of light emitted is:$$\lambda = \frac{1}{4.80	imes10^4 \ m^{-1}} = 208 \ nm$$

Hydrogen Atom: Doubly-Ionized Lithium Is a Hydrogen-Like Ion, in Which