A monochromatic infrared range finder of power $1 ext{ milliwatt}$ emits photons with wavelength $1000 ext

Question

A monochromatic infrared range finder of power $1 	ext{ milliwatt}$ emits photons with wavelength $1000 	ext{ nm}$ in $0.1 	ext{ second}$. The number of photons emitted in $0.1 	ext{ second}$ is: (Given: $h = 6.626 	imes 10^{-34} 	ext{ J s}$, $c = 3 	imes 10^8 	ext{ m s}^{-1}$, Avogadro number $= 6.022 	imes 10^{23}$)

Accepted Answer

1.  **Calculate the Total Energy Emitted ($E_{	ext{total}}$):**
    Power ($P$) is energy per unit time ($P = E/t$).
    Given $P = 1 	ext{ mW} = 1 	imes 10^{-3} 	ext{ J s}^{-1}$ and time $t = 0.1 	ext{ s}$.
    $$E_{	ext{total}} = P 	imes t = (1 	imes 10^{-3} 	ext{ J s}^{-1}) 	imes (0.1 	ext{ s}) = 1 	imes 10^{-4} 	ext{ J}$$

2.  **Calculate the Energy of One Photon ($E_{	ext{photon}}$):**
    Using the Planck equation $E = \frac{hc}{\lambda}$:
    Given $\lambda = 1000 	ext{ nm} = 1000 	imes 10^{-9} 	ext{ m} = 10^{-6} 	ext{ m}$.
    $$E_{	ext{photon}} = \frac{(6.626 	imes 10^{-34} 	ext{ J s})(3 	imes 10^8 	ext{ m s}^{-1})}{10^{-6} 	ext{ m}}$$
    $$E_{	ext{photon}} = \frac{19.878 	imes 10^{-26}}{10^{-6}} 	ext{ J} \approx 1.988 	imes 10^{-19} 	ext{ J}$$

3.  **Calculate the Number of Photons ($N$):**
    $$N = \frac{E_{	ext{total}}}{E_{	ext{photon}}} = \frac{1 	imes 10^{-4} 	ext{ J}}{1.988 	imes 10^{-19} 	ext{ J}}$$
    $$N \approx 0.503 	imes 10^{15} = 5.03 	imes 10^{14}$$

The value is closest to $5 	imes 10^{14}$.

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