1. 686 nm
The lowest photon energy is
[tex]E=1.8 eV[/tex]
Let's convert this energy into Joules first
[tex]E=(1.8 eV)(1.6\cdot 10^{-19} J/eV)=2.9\cdot 10^{-19} J[/tex]
The energy of the photon is given by
[tex]E=\frac{hc}{\lambda}[/tex]
where
h is the Planck constant
c is the speed of light
[tex]\lambda[/tex] is the wavelength
Re-arranging the equation for [tex]\lambda[/tex], we find the maximum wavelength of the photon that can cause a transition:
[tex]\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2.9\cdot 10^{-19} J}=6.86\cdot 10^{-7}m=686 nm[/tex]
2. Visible light
The photon of this light is in the visible light part of the spectrum.
In fact, the range of wavelengths of the visible part of the spectrum is
[380 nm - 750 nm]
In particular, we have that the wavelengths in the range
[640 nm - 750 nm]
corresponds to the red light part of the spectrum: since 686 nm falls withing this range, this photon is a red light photon.
