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

emitted intensity is 7.48 10⁸ 8 W / m²

Step-by-step 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