Assuming that your surface temperature is 990 F and that you

Assuming that your surface temperature is 99.0 F and that you are an ideal blackbody radiator (you are close), find (a) the wavelength at which your spectral radiancy is maximum, (b) the power at which you emit thermal radiation in a wavelength range of 0.950 nm at that wavelength, from a surface area of 4.20 cm², and (c) the corresponding rate at which you emit photons from that area. Using a wavelength of 500 nm (in the visible range), (d) recalculate the power and (e) the rate of photon emission. (As you have noticed, you do not visibly glow in the dark.)

Solution

part a )

T = 99F = 310.372 K

use wien law

lambda_max = 2898 * um .K/T = 2898/310.372 = 9.335 um = 9.335 x 10^-6 m

part b )

lambda = 9.335 um , T = 309.11 K

spectral radiancy = s = (2*pi*c^2*h/lambda^5) * ( 1/e^hc/lambda*k*T -1)

k =boltzman constant

s = 3.58 x 10^7 W/m^3

For small range of wavelength, the radiated power

P = s*A*(dlambda)

dlambda = 0.950 x 10^-9 m

A = 4.20 cm^2 = 4.2 x 10^-4

P =   3.58*10^7*4.2*10^-4* 0.950*10^-9 = 0.0000142=   1.42 x 10^-5 W

part c )

E = hc/lambda

lambda = 9.335 x 10^-6 m

E = 2.2216 x 10^-20 J

dN/dt = P/E = 6.0567 x 10^14 photon/s

part d )

spectral radiancy = s = (2*pi*c^2*h/lambda^5) * ( 1/e^hc/lambda*k*T -1)

lambda = 500 nm = 500 x 10^-9 m

s = 6.255 x 10^-26 W/m^3

P = s*A*dlambda

P = 2.244 x 10^-38 W

part e )

E = hc/lambda

dN/dt = P/E = 5.529 x 10^-20 photon/s