$A$ $25 \ W$ bulb emits monochromatic yellow light of wavelength $0.57 \ \mu m$. Calculate the rate of emission of quanta per second.

Power of bulb,$P = 25 \ W = 25 \ J \ s^{-1}$.
Energy of one photon,$E = \frac{hc}{\lambda}$.
Substituting the values: $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3 \times 10^8 \ m \ s^{-1}$,$\lambda = 0.57 \times 10^{-6} \ m$.
$E = \frac{(6.626 \times 10^{-34} \ J \ s)(3 \times 10^8 \ m \ s^{-1})}{0.57 \times 10^{-6} \ m} = 3.487 \times 10^{-19} \ J$.
Rate of emission of quanta per second = $\frac{P}{E} = \frac{25 \ J \ s^{-1}}{3.487 \times 10^{-19} \ J} = 7.17 \times 10^{19} \ s^{-1}$.