Question

the intensity of a light beam with a wavelength of 620 nm is $4 \times 10^3$ w/m². what is the approximate photon flux?

• $5.47 \times 10^{20}$ m⁻²·s⁻¹
• $8.92 \times 10^{18}$ m⁻²·s⁻¹
• $1.25 \times 10^{22}$ m⁻²·s⁻¹
• $4.51 \times 10^{22}$ m⁻²·s⁻¹
• $3.35 \times 10^{20}$ m⁻²·s⁻¹

Explanation:

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. First, convert the wavelength \( \lambda = 620\, \text{nm}=620\times 10^{-9}\, \text{m} \).

Step2: Calculate photon energy

Substitute the values into the formula:

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Step3: Relate intensity to photon flux

Intensity \( I \) (power per unit area) is the total energy per unit area per unit time. If \( \Phi \) is the photon flux (number of photons per unit area per unit time), then \( I = \Phi E \). So, \( \Phi=\frac{I}{E} \). Given \( I = 4\times 10^{3}\, \text{W/m}^2 = 4\times 10^{3}\, \text{J/(m}^2\text{·s)} \).

Step4: Calculate photon flux

Substitute \( I \) and \( E \) into the formula:

$$
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Answer:

\( 1.25\times 10^{22}\, \text{m}^{-2}\text{·s}^{-1} \) (corresponding to the option "1.25×10²² m⁻²·s⁻¹")