Ans : (d)
$E=hv = \large\frac{hc}{\lambda} = \large\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{ 4560 \times 10^{10}J}$
$= 4.34 \times 10^{-19} J$
$1\: mW$ of light energy is equivalent to $\large\frac{ 10^{-3}}{(4.34 \times 10^{-19} )} = 2.3 \times 10^{15}$ photons/s
The quantum efficiency is $0.5 \%$
This means that only $0.5\: \%$ of these photons release photoelectrons.
So, the number of electrons released from the surface per second
=$ 2.3 \times 10^{15} \times 0.5/100$
$= 1.15 \times 10^{13 }$ electrons/s
The electron current = $1.15 \times 10^{13} \times 1.6 \times 10^{-19} A = 1.84 \: \mu A$