Find the emitted power per square meter and wavelength of peak intensity for a 3000 K object that emits thermal radiation.

power per square meter = 4.593 × 10^(6) W/m²

Wavelength of peak intensity = 9.67 × 10^(-7) m

Explanation:

From Stefan-Boltzmann law, total emitted power per square meter is given as;

P/A = eσT⁴

where;

P is power

A is surface area

σ = Stefan-Boltzmann constant = 5.67 × 10^(-8) W/m².k⁴

T = temperature of the body = 3000 K

e = emissivity of the substance (for ideal radiation, it has a value = 1)

Thus, Plugging in the relevant values we have;

P/A = 1 × 5.67 × 10^(-8) × (3000)^(4)

P/A = 4.593 × 10^(6) W/m²

Let's find the wavelength of peak intensity.

From wiens displacement law, we know that;

λ_m × T = b

where;

λ_m = maximum wavelength

T = Temperature

b is Wien's displacement constant = 2.9 × 10^(−3) m/K

thus;

λ_m = b/T = (2.9 × 10^(−3))/3000 = 9.67 × 10^(-7) m

yeet youwelcome broo

1 a

2 b

3 b

4 d

5 d

6 d

explanation: