The tungsten filament of an incandescent light bulb has a temperature of approximately $3000K$. The emissivity of tungsten is approximately $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$, and you may assume that it is independent of wavelength.

(a) If the bulb gives off a total of $100$watts, 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, between$400and700nm$.

(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 $3695K$.

Step by step solution

01

Part(a): Step 1: Given information

We have been given that the total watts given off by a bulb are$100.$

02

Part(a): Step 2: Solution

Irradiated power is given by$P=Ae\sigma T^4$

Rearranging$A=\frac{P}{e\sigma T^4}$

Putting in the given values and universal constants $=\frac{100\mathrm{W}}{\left(\frac{1}{3}\right)\left(5.67\times {10}^{-8}\mathrm{W}/{\mathrm{m}}^2{\mathrm{K}}^4\right)(3000\mathrm{K}{)}^4}$

$$\begin{gathered} A=6.53\times {10}^{-5}{\mathrm{m}}^2 \\ A=6.53\times {10}^{-5}{\times {10}^6\mathrm{mm}}^2 \\ A=65.3{\mathrm{mm}}^2 \end{gathered}$$

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 by$u\left(T\right)=\frac{8\pi }{(hc{)}^3}\frac{{ϵ}^3}{e^{ϵ/kT}-1}$

let$x=ϵ/kT$

$u\left(T\right)=\frac{8\pi (kT{)}^3}{(hc{)}^3}\frac{x^3}{e^x-1}$

For peak, partial derivative = 0

$\frac{\partial }{\partial x}\left(\frac{x^3}{e^x-1}\right)=0$

$\frac{3x^2\left(e^x-1\right)-x^3{e}^x}{{\left(e^x-1\right)}^2}=0$

$3x^2\left(e^x-1\right)-x^3{e}^x=0$

$3e^x-3-x{e}^x=0$

$x=2.8214$

Also$ϵ=xkT$

$ϵ=(2.8214)\left(13.8\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)(3000\mathrm{K})$

$ϵ=1.168\times {10}^{-18}\mathrm{J}$

Wavelength $\mathrm{ϵ}=\mathrm{hc}/\mathrm{\lambda }$

$\lambda =\frac{hc}{ϵ}$

$\lambda =1.70\times {10}^{-7}\mathrm{m}$

05

Part(c) Step 1:Given information

We have been given that${ϵ}_{400\mathrm{nm}}=\frac{\left(6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)\left(3.0\times {10}^8\mathrm{m}/\mathrm{s}\right)}{400\times {10}^{-9}\mathrm{m}}=4.97\times {10}^{-19}=3.1\mathrm{eV}$

06

Part(c) Step 2:Simplify

The graph of this part as shown

07

Part(d) Step 1:Given information

We have been given that$f=\frac{U_{vis}}{U_{tot}}$

08

Part(d) Step 2:Simplify

The integration of the range of visible energy is

$$\begin{gathered} U\propto \int \frac{{ϵ}^3}{e^{ϵ/kT}-1}dϵ \\ {U}_{vis}\propto {\int }_{vis}\frac{x^3}{e^x-1}dx \end{gathered}$$

The boundaries are

$x_2=\frac{{ϵ}_{400\mathrm{nm}}}{kT}=\frac{3.1\mathrm{eV}}{\left(8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)(3000\mathrm{K})}=11.987$

$x_1=\frac{{ϵ}_{700\mathrm{nm}}}{kT}=\frac{1.77\mathrm{eV}}{\left(8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)(3000\mathrm{K})}=6.845$

09

Part(e) Step 1: Given information

We have been given that$f=\frac{15}{{\pi }^4}{\int }_{x_1}^{x_2}\frac{x^3}{e^x-1}dx$

10

Part(e) Step 2: Simplify

As the temperature increases,the efficiency increases

$x_2=\frac{{ϵ}_{400\mathrm{nm}}}{kT}{x}_1=\frac{{ϵ}_{700\mathrm{nm}}}{kT}$

11

Part(f) Step 1:Given information

We have been given that$x_2=\frac{7}{4}{x}_1$

12

Part(f) Step 2:Simplify

The tempreature at this point is:

$T=\frac{{ϵ}_{700\mathrm{nm}}}{k{x}_1}=\frac{1.77\mathrm{eV}}{\left(8.62\times {10}^{-5}\mathrm{eV}/\mathrm{K}\right)(2.92)}=7032\mathrm{K}$

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