Question

The single electron on a Li2+(g) atom is excited to the n=7 orbital. When the electron falls to the n=3 orbital a photon of light is emitted by the atom. What is the wavelength of this photon of light?

(a) 496 nm

(b) 1005 nm

(c) 112 nm

(d) 91.2 nm

(e) 4468 nm

** solutions says answer is (C) but not sure how they got that

Answer 1

let the wavelength be x.so,

the wavelength is given by,

1/x=10967700*Z^2*(1/n1^2 - 1/n2^2)

or 1/x=10967700*3^2*(1/3^2 - 1/7^2)

or x=111.692 nm

~ 112 nm