Question

Question9:                                                                                                                     

The tungsten light bulb filament operates at about 3000K.

1. )     Determine the median wavelength emitted by the light bulb.
2.    I identify the region in the electromagnetic spectrum this wavelength lies.
3. If it’s a100-W bulb, calculate is the filament’s surface area.

Answer 1

Solution:-

a.)

Median length emitted by light bulb

λ = b/T
b = wein's displacement constant
= 2.897*10^-3 m.k
T = 3000 k
So λ = (2.897*10^-3)/3000 m
λ = 966 nm
Wavelength = 966 nm

b.) For range 750 nm to 1mm wavelength

Infra red(IR) electromagnetic spectrum is seen as the region where wavelength lies.

c.) Use Stefan-BOltzman law:

P/A = esT^4

P = power radiated in watt, A = area of radiator, e= emissivity of radiator, s = 5.67x10^-8 W/m^2-K^4, and T = temperature in Kelvin.

Here emissivity of radiator =1/3 (Assumed)

Now after solving we get

So A = 100/(0.333*5.67x10^-8*(3000)^4) = 6.53*10^-5 m^2

Filament's surface area (A) = 6.53*10^-5 m^2