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

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

Step-by-step 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