Answer:
2πc/w
Explanation:
To find the wavelength you take into account the difference in energy of two adjacent states n+1 and n:
[tex]E_{n+1,n}=\hbar \omega((n+1)+\frac{1}{2})+\hbar \omega(n+\frac{1}{2})\\\\E_{n+1,n}=\hbar \omega[/tex](1)
hbar = h/2π
this energy is also the energy of an emitted photon in the transition, that is:
[tex]E_{\lambda}=h\frac{c}{\lambda}[/tex] (2)
you equal the equations (1) and (2) and compute the wavelength:
[tex]E_{\lambda}=E_{n+1,n}\\\\h\frac{c}{\lambda}=\frac{h}{2\pi}\omega\\\\\lambda=\frac{2\pi c}{\omega}[/tex]
hence, the wavelength of the emitted photon is 2πc/w
