Question

A watt is a unit of energy per unit time, and one watt (W) is equal to one joule per second ( J ⋅ s − 1 ) J⋅s−1) . A 40.0 W 40.0 W incandescent lightbulb produces about 5.00 % 5.00% of its energy as visible light. Assuming that the light has an average wavelength of 510.0 nm, calculate how many such photons are emitted per second by a 40.0 W 40.0 W incandescent lightbulb.

Answer 1

Answer:

5.12 x 10^18 photon per second

Explanation:

Power of bulb, P = 40 W

%5 of this energy is visible, so the visible energy = E = 5% of 40 = 2 J/s

Wavelength of light, λ = 510 nm = 510 x 10^-9 m

let the number of photons per second is n.

Energy of each photon,

E' = hc / λ

$E' = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{510\times 10^{-9}}$

E' = 3.9 x 10^-19 J

number of photons per second, n = E / E'

$n=\frac{2}{3.9\times 10^{-19}}$

n = 5.12 x 10^18 per second

So, the number of photons per second are 5.12 x 10^18.