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  • A light bulb emits photons of various wavelengths ranging from 400 nm to 700 nm. If the power of the light bulb is 100 W, what is the total number of photons emitted per second?

A light bulb emits photons of various wavelengths ranging from 400 nm to 700 nm. If the power of the light bulb is 100 W, what is the total number of photons emitted per second?

Option: 1

1.78 \times 10^{20}$ photons $/ \mathrm{s}


Option: 2

2.01 \times 10^{20}$ photons $/ \mathrm{s}


Option: 3

2.31 \times 10^{20}$ photons $/ \mathrm{s}


Option: 4

2.77 \times 10^{20}$ photons $/ \mathrm{sec}


Answers (1)

best_answer

The energy of each photon can be calculated using the formula E=\frac{h c}{\lambda}, where \mathrm{h} is Planck's constant \left(6.6 \times 10^{-34} \mathrm{Js}\right) \mathrm{c} is the speed of light \left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right).

For a range of wavelengths from 400 \mathrm{~nm}$ to $700 \mathrm{~nm}, the average wavelength can be calculated as \lambda_{a v g}=\frac{400+700}{2}=550 \mathrm{~nm}.
Using the formula for the energy of each photon, we get:
$$ \begin{gathered} E=\frac{h c}{\lambda_{a v g}}=\frac{\left(6.6 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)}{550 \times 10^{-9} \mathrm{~m}}= 3.61 \times 10^{-19} J \end{gathered}


The power of the light bulb is given as P=100 \mathrm{~W}.


Using the formula for the number of photons emitted per second, we get:
$$ \begin{aligned} & n=\frac{P}{E}=\frac{100}{3.61 \times 10^{-19}}=2.77 \times 10^{20} \text { photons } / s \end{aligned}
Therefore, the total number of photons emitted per second is 2.77 \times 10^{20} \mathrm{photons} / \mathrm{s}.

Hence, the correct answer is an option (4).


 

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