Monochromatic light of wavelength 3000 Å is incident on a surface area 4 cm². If intensity of light is 150 mW/m², then rate at which photons strike the target is

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9 x 10¹³/sec

7 x 10¹⁵/sec

3 x 10¹⁰/sec

6 x 10¹⁹/sec

answer is B.

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Detailed Solution

Calculation of Photon Flux Incident on a Surface

Given monochromatic light of wavelength 3000 Å, incident on a surface area of 4 cm². The intensity of the light is 150 mW/m².

The rate at which photons strike the target is given by:

n / t = (I × A) / (hc / λ)

Where:
I = intensity = 150 × 10^-3 W/m²
A = surface area = 4 × 10^-4 m²
h = Planck's constant = 6.6 × 10^-34 J·s
c = speed of light = 3 × 10⁸ m/s
λ = wavelength = 3 × 10^-7 m (3000 Å)

Calculating photon energy:

E_photon = hc/λ = (6.6 × 10^-34 × 3 × 10⁸) / 3 × 10^-7 = 6.6 × 10^-19 J

Power incident on surface:

P = I × A = 150 × 10^-3 × 4 × 10^-4 = 6 × 10^-5 W

Photon flux (number of photons per second):

n/t = P / E_photon = (6 × 10^-5 ) / (6.6 × 10^-19) = 9 × 10¹³ photons/s

Hence, around 9 × 10¹³ photons strike the target per second.

$\frac{n}{t} = \frac{{IA\lambda }}{{hc}} = \frac{{150 \times {{10}^{ - 3}} \times 4 \times {{10}^{ - 4}} \times 3 \times {{10}^{ - 7}}}}{{6.6 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}} = 9 \times {10^{13}}\frac{1}{{sec}}$