Photons Emitted per Second – Rankers Physics
Topic: Modern Physics
Subtopic: Photoelectric Effects and deBroglie Equation

Photons Emitted per Second

The number of photons per second on an average emitted by the source of monochromatic light of wavelength \(600\text{ nm}\), when it delivers the power of \(3.3 \times 10^{-3}\text{ watt}\) will be (\(h = 6.6 \times 10^{-34}\text{ J s}\))
\(10^{15}\)
\(10^{18}\)
\(10^{17}\)
\(10^{16}\)

Solution:

Power \(P = n \frac{hc}{\lambda}\), where \(n\) is the number of photons emitted per second. Substituting the given values, we get \(n = \frac{3.3 \times 10^{-3} \times 600 \times 10^{-9}}{6.6 \times 10^{-34} \times 3 \times 10^8} = 10^{16}\text{ s}^{-1}\).

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