Question

The energy of the photons corresponding to the peak emission of a hot body is equal to 4 eV. Calculate the total emitted intensity from that body, in units of W/m^2.

Answer 1

Answer:

emitted intensity is 7.48 10⁸ 8 W / m²

Explanation:

To calculate this energy, let's start by using the win displacement law, and the Planck equation

       λ T = 2,898 10-3 m K

       E = h f

       c = f λ

       E = h c /λ

       λ = hc / E

calculate

       (h c / E) T = 2,898 10⁻³

       T = 2,898 10⁻³ E / h c

       E = 4 eV (1.6 10⁻¹⁹ J / 1eV) = 6.4 10⁻¹⁹ J

       T = 2,898 10⁻³  6.4 10⁻¹⁹ / (6,626 10⁻³⁴ 3 10⁸)

       T = 18.54 10⁻²² / (19.878 10⁻²⁶)

       T = 0.9331 10⁴ K

Having the temperature at which the emission occurs we can use Stefan's law

          P = σ A e T⁴

The constant σ = 5.670 10⁻⁸ W / m² K⁴, e is the emissivity for a black body worth 1

          P / A = 5,670 10⁻⁸ 1 (0.9331 10⁴)⁴

          P / A = 7.48 10⁸ 8 W / m²