Using the bohr model, determine the energy in joules of the photon produced when an electron in a li2+ ion moves from the orbit with n = 2 to the orbit with n = 1.

1.472 x 10^-17 J energy of the photon is released (exothermic).

Explanation:

E = –kZ^2 / n^2

k = 2.18 x10^-18 J

Z = atomic number

Li2+ atomic number = 3

Ei = –(2.18 x 10^-18 J) x 3^2 / 2^2 = –4.905 x 10^-18 J

Ef = –(2.18 x 10^-18 J) x 3^2 / 1^2 = –1.962 x 10^-17 J

ΔE = |Ef – Ei| = |(–1.962 x 10^-17 J) – (–4.905 x 10^-18 J)| = 1.472x10^-17  

1.472 x 10-17 J energy of the photon is released (exothermic).

im pretty sure its b