Concept:

Total light power output is quantized into discrete photons. The number of photons emitted per second is the total visible power divided by the energy of a single photon.

Formula:

$$ P_{light} = P_{total} \times \text{Efficiency} $$
$$ E_{photon} = \frac{hc}{\lambda} $$
$$ n = \frac{P_{light}}{E_{photon}} $$

Solution:

• Effective light power \( P = 5 \text{ W} \times 0.80 = 4 \text{ J/s} \).



• Energy per photon \( E = \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{660 \times 10^{-9}} = \frac{19.8 \times 10^{-26}}{6.6 \times 10^{-7}} = 3 \times 10^{-19} \text{ J} \).



• Number of photons \( n = \frac{4}{3 \times 10^{-19}} \approx 1.33 \times 10^{19} \text{ photons/s} \).

Why other options are incorrect:

Forgetting to account for the 80% efficiency factor gives \( 1.6 \times 10^{19} \). Incorrect unit conversions for nanometers (using \( 10^{-6} \) instead of \( 10^{-9} \)) leads to off-by-magnitude errors.