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 incandesc

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.

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Sapo 4 years 2021-07-18T23:54:52+00:00 1 Answers 82 views 0

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    2021-07-18T23:56:33+00:00

    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.

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