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? A) 16.9 µm B) 1.88 µm C) 470 nm D) 208 nm

Question

Quizplus · Accepted Answer

The energy difference between n=4 and n=3 orbits is given by the formula:
ΔE = 13.6 * Z^2 * (1/nf^2 - 1/ni^2)
where Z = 3 (charge of the nucleus), nf = 3 (final orbit), and ni = 4 (initial orbit).
Plugging in these values, we get:
ΔE = 13.6 * 3^2 * (1/3^2 - 1/4^2) = 5.925 eV
We can use the formula ΔE = hc/λ to find the wavelength of the emitted light, where h is Planck's constant, c is the speed of light, and λ is the wavelength.
Rearranging the formula, we get:
λ = hc/ΔE
Substituting the values, we get:
λ = (6.626 x 10^-34 J.s) * (2.998 x 10^8 m/s) / (5.925 eV * 1.602 x 10^-19 J/eV) = 208 nm
Therefore, the answer is D) 208 nm.

Doubly-Ionized Lithium Is a Hydrogen-Like Ion,in Which the Nucleus with Charge