A Li²⁺ ion undergoes a transition from the n = 4 to the n = 3 state. The energy of the emitted photon is A) 4.5 eV B) 10.2 eV C) 5.95 eV D) 2.6 eV E) 0.66 eV

Question

Quizplus · Accepted Answer

The energy difference between two energy levels, n1 and n2, in an atom is given by the formula:
ΔE = hc/λ = E(n2) - E(n1) = (13.6 eV) [(1/n1²) - (1/n2²)]
where h is Planck's constant, c is the speed of light, λ is the wavelength of the emitted photon, and E(n) is the energy of the nth energy level of the hydrogen atom.

For a Li2+ ion, which has only one electron, we can use the same formula, but with a modified value of the atomic number Z:
ΔE = (13.6 eV) [(1/n1²) - (1/n2²)] Z²

In this case, n1 = 4 and n2 = 3, so we have:
ΔE = (13.6 eV) [(1/4²) - (1/3²)] (3)² = 5.95 eV

Therefore, the energy of the emitted photon is 5.95 eV. The correct choice is C.

A Li2+ Ion Undergoes a Transition from the N =

A Li²⁺ Ion Undergoes a Transition from the N =