Answer:
1.88 × 10^21 photons per second
Step-by-step explanation:
The number of photons emitted per second by a blackbody can be calculated using the Stefan-Boltzmann law and Planck's law. The Stefan-Boltzmann law gives the total power radiated by a blackbody in terms of its temperature T:
P = σ * A * T^4
where σ is the Stefan-Boltzmann constant (σ = 5.67 × 10^-8 W/m^2 K^4), A is the surface area of the blackbody, and T is its temperature.
Planck's law gives the spectral radiance of a blackbody at a given wavelength and temperature:
B(λ, T) = (2hc^2 / λ^5) * (1 / (exp(hc / λkT) - 1))
where B is the spectral radiance, h is Planck's constant (h = 6.626 × 10^-34 J s), c is the speed of light, k is Boltzmann's constant (k = 1.38 × 10^-23 J/K), λ is the wavelength, and T is the temperature.
The wavelength of peak intensity (λ_max) can be found using Wien's displacement law:
λ_max = b / T
where b is the Wien displacement constant (b ≈ 2.898 × 10^-3 m K).
Assuming that all photons are emitted at the wavelength of peak intensity, the spectral radiance at λ_max can be found by substituting λ = λ_max into Planck's law:
B(λ_max, T) = (2hc^2 / λ_max^5) * (1 / (exp(hc / (λ_max kT)) - 1))
The total number of photons emitted per second by the blackbody can be found by dividing the total power radiated by the energy per photon at the wavelength of peak intensity:
N = P / (hc / λ_max)
Substituting the given values (assuming a blackbody with unit surface area), we get:
λ_max = b / T = (2.898 × 10^-3 m K) / 300 K = 9.66 × 10^-6 m
B(λ_max, T) = (2hc^2 / λ_max^5) * (1 / (exp(hc / (λ_max kT)) - 1)) ≈ 1.19 × 10^-12 W/m^2/m
P = σ * A * T^4 = (5.67 × 10^-8 W/m^2 K^4) * (1 m^2) * (300 K)^4 = 384.6 W
hc / λ_max = (6.626 × 10^-34 J s) * (3 × 10^8 m/s) / (9.66 × 10^-6 m) = 2.046 × 10^-19 J
N = P / (hc / λ_max) ≈ (384.6 W) / (2.046 × 10^-19 J) ≈ 1.88 × 10^21 photons/s
Therefore, if we make the approximation that all photons are emitted at the wavelength of peak intensity, the body emits approximately 1.88 × 10^21 photons per second.