Answer:
3 photons
Explanation:
The energy of a photon E can be calculated using this formula:
[tex]E=\frac{hc}{\lambda}[/tex]
Where [tex]h[/tex] corresponds to Plank constant (6.626070x10^-34Js), [tex]c[/tex] is the speed of light in the vacuum (299792458m/s) and [tex]\lambda[/tex] is the wavelength of the photon(in this case 800nm).
[tex]E=\frac{hc}{\lambda}=\frac{(6.626070\times10^{-34})(299792458)}{800\times10^{-9}}=\frac{1.986445812\times10^-25}{800}=2.483057265\times10^{-19}J[/tex]
Tranform the units
[tex]1eV=1.602176634\times10^{-19}J\\2.483057265\times10^{-19}J(\frac{1eV}{1.602176634\times10^{-19}J})=1.549802445eV[/tex]
The band Gap is 4eV, divide the band gap between the energy of the photon:
[tex]\frac{4ev}{1.549802445eV}=2.508974118[/tex]
Rounding to the next integrer: 3.
Three photons are the minimum to equal or exceed the band gap.