Answer
$3.26 \times10^{18}$ $s^{-1}$
Work Step by Step
Let $R_{1}$ be the number of photons emitted per unit time (emission rate) and $E$ is the energy of per photon.
$\therefore$ The output power of the Argon laser is given by
$P=R_{1}E$
or, $R_{1}=\frac{P}{E}$
Energy of a photon is expresses as, $E=\frac{ch}{\lambda}$,
in which $c$ is the speed of light in vacuum, $\lambda$ is the wavelength of the light and $h$ is the Planck's constant.
$\therefore$ $R_{1}=\frac{P}{\frac{ch}{\lambda}}=\frac{P\lambda}{ch}$
or, $R_{1}=\frac{1.5\times 515\times 10^{-9}}{3\times 10^{8} \times 6.63\times 10^{-34}}$ $s^{-1}$
or, $R_{1}\approx 3.88 \times 10^{18}$ $s^{-1}$
It is given that 84% of the incident energy ends up within this central disk.
Therefore, the rate are photons absorbed by the screen in the central disk of the diffraction pattern is given by
$84\%$ of $R_{1}$
$=0.84\times 3.88 \times 10^{18}$ $s^{-1}$
$\approx 3.26 \times10^{18}$ $s^{-1}$
