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The tungsten filament of an incandescent light bulb has a temperature of approximately 3000K. The emissivity of tungsten is approximately 13, and you may assume that it is independent of wavelength.

(a) If the bulb gives off a total of 100watts, what is the surface area of its filament in square millimetres?

(b) At what value of the photon energy does the peak in the bulb's spectrum occur? What is the wavelength corresponding to this photon energy?

(c) Sketch (or use a computer to plot) the spectrum of light given off by the filament. Indicate the region on the graph that corresponds to visible wavelengths, between400and700nm.

(d) Calculate the fraction of the bulb's energy that comes out as visible light. (Do the integral numerically on a calculator or computer.) Check your result qualitatively from the graph of part (c).

( e) To increase the efficiency of an incandescent bulb, would you want to raise or lower the temperature? (Some incandescent bulbs do attain slightly higher efficiency by using a different temperature.)

(f) Estimate the maximum possible efficiency (i.e., fraction of energy in the visible spectrum) of an incandescent bulb, and the corresponding filament temperature. Neglect the fact that tungsten melts at 3695K.

Short Answer

Expert verified

(a) The surface area of its filament is A=65.3mm2.

(b) The wavelength corresponding to this photon energy is=1.7010-7m.

(c) The spectrum of the light given off is 1.7010-7m

(d) The fraction of the bulb's energy that comes out as visible light isf=0.08124.

(e) To increase the efficiency of the incandescent bulb we will raise the temperature.

(f) The maximum possible efficiency is achieved at temperatureT=7032K.

Step by step solution

01

Part(a): Step 1: Given information

We have been given that the total watts given off by a bulb are100.

02

Part(a): Step 2: Solution

Irradiated power is given byP=AeT4

RearrangingA=PeT4

Putting in the given values and universal constants =100W135.6710-8W/m2K4(3000K)4

A=6.5310-5m2A=6.5310-5106mm2A=65.3mm2

03

Part(b): Step 1: Given information

We have to find out at what value of the photon energy does the peak in the bulb's spectrum occur

04

Part(b): Step 2: Solution

Spectrum is given byu(T)=8(hc)33e/kT-1

letx=/kT

u(T)=8(kT)3(hc)3x3ex-1

For peak, partial derivative = 0

xx3ex-1=0

3x2ex-1-x3exex-12=0

3x2ex-1-x3ex=0

3ex-3-xex=0

x=2.8214

Also=xkT

=(2.8214)13.810-23J/K(3000K)

=1.16810-18J

Wavelength =hc/

=hc

=1.7010-7m

05

Part(c) Step 1:Given information

We have been given that400nm=6.62610-34Js3.0108m/s40010-9m=4.9710-19=3.1eV

06

Part(c) Step 2:Simplify

The graph of this part as shown

07

Part(d) Step 1:Given information

We have been given thatf=UvisUtot

08

Part(d) Step 2:Simplify

The integration of the range of visible energy is

U3e/kT-1dUvisvisx3ex-1dx

The boundaries are

x2=400nmkT=3.1eV8.6210-5eV/K(3000K)=11.987

x1=700nmkT=1.77eV8.6210-5eV/K(3000K)=6.845

09

Part(e) Step 1: Given information

We have been given thatf=154x1x2x3ex-1dx

10

Part(e) Step 2: Simplify

As the temperature increases,the efficiency increases

x2=400nmkTx1=700nmkT

11

Part(f) Step 1:Given information

We have been given thatx2=74x1

12

Part(f) Step 2:Simplify

The tempreature at this point is:

T=700nmkx1=1.77eV8.6210-5eV/K(2.92)=7032K

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