Use the Planck expression to get the energy of 1 photon:
#sf(E=hf)#
#sf(f)# is the frequency
#sf(h)# is the Planck Constant = #sf(6.63xx10^(-34)color(white)(x)J.s)#
Since we are given the wavelength #sf(lambda)# this becomes:
#sf(E=(hc)/(lambda))#
Where #sf(c)# is the speed of light which I will take to be #sf(3.0xx10^(8)color(white)(x)"m/s")#.
#:.##sf(E=(6.63xx10^(-34)xx3.0xx10^(8))/(700xx10^(-9))=2.841xx10^(-19)color(white)(x)J)#
1 Watt is 1 Joule of work done per second.
#sf(0.1color(white)(x)J)# of work is being done in 1 second.
So
#sf(2.841xx10^(-19)color(white)(x)J)# is produced from 1 photon.
#:.##sf(1color(white)(x)J)# is produced from #sf(1/(2.841xx10^(-19)))# photons.
#:.##sf(0.1color(white)(x)J)# is produced from #sf(1/(2.841xx10^(-19))xx0.1=3.5xx10^(17))# photons in 1 second.
