Question

A 100-watt light bulb radiates energy at a rate of 100 J/s. (The watt, a unit of power or energy over time, is defined as 1 J/s J/s.) If all of the light emitted has a wavelength of 525nm 525nm, how many photons are emitted per second? Express your answer to three significant figures.

Answer 1

Answer:

Approximately $2.64\times 10^{20}$ (assumption: wavelength $525\; \rm nm$ is measured in vacuum, where the speed of light is approximately $3.0\times 10^{8}\; \rm m \cdot s^{-1}$.)

Explanation:

Convert the unit of wavelength to meters:

$\displaystyle \lambda = 525\; \rm nm = 525 \; \rm nm\times \frac{10^{-9}\; \rm m}{1\; \rm nm} = 5.25 \times 10^{-7}\; \rm m$.

Assume that the wavelength $525\; \rm nm$ is measured in vacuum, where the speed of light is approximately $2.99792\times 10^{8}\; \rm m \cdot s^{-1}$. Calculate the frequency of this light from its wavelength:

$\displaystyle f = \frac{c}{\lambda} \approx \frac{2.99792\times 10^{8}\; \rm m \cdot s^{-1}}{5.25 \times 10^{-7}\; \rm m} \approx 5.71429\times 10^{14}\; \rm s^{-1}$.

The Planck's Constant can help find the energy of a photon given its frequency. Look up this constant to more than three significant figures:

$h \approx 6.62607\times 10^{-34}\; \rm J \cdot s^{-1}$.

Calculate the energy of one such photon:

$\begin{aligned} E &= h \cdot f\\ &\approx 6.62607\times 10^{-34}\; \rm J \cdot s^{-1} \times 5.71023\times 10^{14}\; \rm s \\ &\approx 3.78370\times 10^{-19}\; \rm J \end{aligned}$.

Calculate the number of these photons that $100\; \rm J$ of energy can produce under the assumption of $100\%$ conversion:

$\displaystyle \frac{100\; \rm J}{3.78370\times 10^{-19}\; \rm J} \approx 2.64\times 10^{20}$.

(Rounded to three significant figures.)