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Q.
A 150 watt bulb emits light of wavelength 6600 $mathring{A}$ and only 8% of the energy is emitted as light. How many photons are emitted by the bulb per second?
Power of the bulb = $150\, watt = 150 \, J\, s^{-1}$
As only 8% of the energy is emitted as light so, the total energy emitted per second.
$ = \frac{150 \, J \times 8}{100} = 12 \, J$
Energy of one photon, $E = hv = \frac{hc}{\lambda}$
$= \frac{\left(6.626 \times10^{-34} J s\right) \times \left(3\times 10^{8} m s^{-1}\right)}{6600 \times 10^{-10} m} $
$= 3.0118 \times 10^{-19} J $
$\therefore $ Number of photons emitted
$ = \frac{12J}{3.0118 \times 10^{-19} J} = 3.98 \times 10^{19} \approx4.0 \times 10^{19 }$