Question

A 5 watt LED bulb converts 80% of the power into light photons of wavelength 660nm. What is the number of photons emitted from the bulb in one second ?​

Answer 1

We determine the power of all photons

$P_{\sum ph}=0.8\cdot P_{LED}=0.8\cdot 5=4watt$

Also, the power of all photons is equal to

$P_{\sum ph}=P_f\cdot N_{f}$

where will we write

$N_f=\frac{P_{\sum ph}}{P_f}=\frac{P_{\sum ph}}{(h_ \cdot \nu)/t}=\frac{P_{\sum ph}}{(h_ \cdot \frac{с}{\lambda})/t}=\frac{t\cdot\lambda\cdot P_{\sum ph}}{h \cdot c}$

Substitute the numbers, we get

$N_f=\frac{1(s)\cdot660 \cdot10^{-9}(m) \cdot 4(watt)}{6.626\cdot10^{-34}(J\cdot s) \cdot 3 \cdot 10^8 (m/s)}=1.328 \cdot 10^{19} photons$