QUESTION IMAGE
Question
a 150 w lamp emits light of wavelength 590 nm uniformly in all directions. what is the photon flux (photons per unit area per unit time) on a small screen at a distance 2.3 m from the lamp?
○ 1.3 × 10¹⁹ photons/m²·s
○ 1.5 × 10¹⁹ photons/m²·s
○ 2.7 × 10¹⁹ photons/m²·s
○ 8.4 × 10¹⁹ photons/m²·s
○ 6.7 × 10¹⁸ photons/m²·s
Step1: Recall photon energy formula
The energy of a single photon is given by \( E = \frac{hc}{\lambda} \), where \( h = 6.626\times 10^{-34}\, \text{J·s} \) (Planck's constant), \( c = 3\times 10^{8}\, \text{m/s} \) (speed of light), and \( \lambda \) is the wavelength of light. First, convert the wavelength \( \lambda = 590\, \text{nm}=590\times 10^{-9}\, \text{m} \).
Step2: Calculate photon energy
Substitute values into the formula: \( E=\frac{6.626\times 10^{-34}\times3\times 10^{8}}{590\times 10^{-9}} \). Calculate numerator: \( 6.626\times 10^{-34}\times3\times 10^{8}=1.9878\times 10^{-25}\, \text{J·m} \). Then divide by \( 590\times 10^{-9}\, \text{m} \): \( E=\frac{1.9878\times 10^{-25}}{590\times 10^{-9}}\approx 3.37\times 10^{-19}\, \text{J} \).
Step3: Find total number of photons per second
The power of the lamp \( P = 150\, \text{W}=150\, \text{J/s} \). The number of photons emitted per second \( n=\frac{P}{E} \). Substitute \( P = 150 \) and \( E\approx 3.37\times 10^{-19} \): \( n=\frac{150}{3.37\times 10^{-19}}\approx 4.45\times 10^{20}\, \text{photons/s} \).
Step4: Calculate photon flux
The lamp emits light uniformly in all directions, so the light spreads over a spherical surface area \( A = 4\pi r^{2} \), where \( r = 2.3\, \text{m} \). Calculate \( A = 4\pi(2.3)^{2}\approx 4\pi\times 5.29\approx 66.47\, \text{m}^2 \). Photon flux \( \Phi=\frac{n}{A} \). Substitute \( n\approx 4.45\times 10^{20} \) and \( A\approx 66.47 \): \( \Phi=\frac{4.45\times 10^{20}}{66.47}\approx 6.7\times 10^{18}\, \text{photons/m}^2\text{·s} \).
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\( 6.7\times 10^{18}\, \text{photons/m}^2\text{·s} \) (corresponding to the option: \( 6.7\times 10^{18}\, \text{photons/m}^2\text{·s} \))