Answer
$5.5\times10^{18}\frac{photons}{s}$
Work Step by Step
The impulse on the wall is equal and opposite to the change in momentum of the photons. The photons are absorbed and transfer all of their momentum to the wall.
Equation 27–6 gives us the momentum of one photon.
$$p_{photon}= \frac{h}{\lambda}$$
Relate the magnitude of the force on the wall to the magnitude of the photons’ change in momentum. Let n represent the number of photons hitting the wall in a time $\Delta t$.
$$F_{wall}\Delta t=n p_{photon}=n \frac{h}{\lambda}$$
Number per second = $\frac{n}{\Delta t}=\frac{ F_{wall}\lambda}{h}$
$$=\frac{(5.8\times10^{-9}N)(633\times10^{-9}m)}{6.63\times10^{-34}J \cdot s }$$
$$=5.5\times10^{18}\frac{photons}{s}$$
