Question

Calculate the temperature of and power-per-area radiated from a blackbody if the spectral intensity density peaks at (a) gamma rays, λ = 1.50×10−14 m; (b) x rays, 1.50 nm; (c) red light, 640 nm; (d) broadcast television waves, 1.00 m; and (e) AM radio waves, 204 m.

Answer 1

Answer:

a)  T = 1,932 10¹¹ K ,  I = 7.90 10³⁷ W / m² ,

b)  T = 1,932 10⁶ K ,  I = 7.90 10¹⁷ W / m² ,

c)  T = 4,528 10³ K ,  I = 4.20 10⁶ W / m² ,

d)  T = 2,898 10⁻³ K ,  I = 4.0 10⁻¹⁰ W / m²

e)   T = 1.42 10⁻⁵ K , I= 4.07 10⁻²⁸ W / m²

Explanation:

For this exercise we can use Wien's displacement law and Stefan's law

          λ T = 2,898 10⁻³

         T = 2,898 10-3 / λ

         P = σ A e T⁴

         I = P / A = σ e T⁴

The value of the Stefan –Boltzmann (σ) constant is 5.670 10⁻⁸ W / m²K⁴, the emissivity (e) for a black body is  1

           I = σ T⁴

a) λ = 1.50 10⁻¹⁴ m

The temperature is

            T = 2,898 10⁻³ /λ

            T = 2,898 10⁻³ / 1.50 10⁻¹⁴

            T = 1,932 10¹¹ K

            I = 5,670 10⁻⁸ (1,932 10¹¹)⁴

            I = 7.90 10³⁷ W / m²

b) λ = 1.50 nm = 1.50 10⁻⁹ m

              T = 2,898 10⁻³ / 1.50 10⁻⁹

              T = 1,932 10⁶ K

              I = 5,670 10⁻⁸ (1,932 10⁶)⁴

              I = 7.90 10¹⁷ W / m²

c) λ = 640 nm = 6.40 10⁻⁷ m

              T = 2,898 10⁻³ / 6.40 10⁻⁷

              T = 4,528 10³ K

              I = 5,670 10⁻⁸ (4,528 10³)⁴

              I = 4.20 10⁶ W / m²

d) λ = 1.00 m

             T = 2,898 10⁻³ / 1

             T = 2,898 10⁻³ K

             I = 5,670 10⁻⁸ (2,898 10⁻³)⁴

             I = 4.0 10⁻¹⁰ W / m²

e) λ = 204 m

            T = 2,898 10⁻³ / 204

            T = 1.42 10⁻⁵ K

             I = 5,670 10⁻⁸ (1.42 10⁻⁵)⁴

             I = 4.07 10⁻²⁸ W / m²