A parallel beam of light of wavelength $900 \,nm$ and intensity $100 \,Wm ^{-2}$ is incident on a surface perpendicular to the beam Tire number of photons crossing $1 \,cm ^2$ area perpendicular to the beam in one second is :
To solve the problem of finding the number of photons crossing a given area per second, we'll follow these steps:
First, we'll calculate the energy of a single photon using the formula:
E = \frac{hc}{\lambda},
where h = 6.626 \times 10^{-34} \, Js (Planck's constant), c = 3 \times 10^8 \, m/s (speed of light), and \lambda = 900 \, nm = 900 \times 10^{-9} \, m (wavelength).
Substituting the values into the formula gives:
E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{900 \times 10^{-9}} \, J
Calculating, we find:
E = \frac{19.878 \times 10^{-26}}{900 \times 10^{-9}} = 2.2087 \times 10^{-19} \, J
Next, we determine the number of photons by dividing the intensity of the light by the energy of a single photon:
The intensity formula is related to the number of photons (N) as:
I = \frac{N \times E}{A \times t},
where A = 1 \, cm^2 = 1 \times 10^{-4} \, m^2 and t = 1 \, s.
Rearranging, we find:
N = \frac{I \times A \times t}{E}
Substituting the given values,
N = \frac{100 \times 1 \times 10^{-4} \times 1}{2.2087 \times 10^{-19}}
Simplifying this, we get:
N = \frac{100 \times 10^{-4}}{2.2087 \times 10^{-19}} = \frac{10^{-2}}{2.2087 \times 10^{-19}}
Solving gives:
N \approx 4.526 \times 10^{16}
↪ which can be approximated as 4.5 \times 10^{16}.
Thus, the correct answer is 4.5 \times 10^{16}, as this matches one of the provided options.
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