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3 May, 23:18

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.

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  1. 3 May, 23:20
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    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
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