Question

Relationships described in Problems $7.3$and $7.4$may be of help in solving the following.

(a) Calculate the wavelength of maximum emission of a tungsten filament bulb operated at the usual temperature of localid="1646640900138" $2870\mathrm{K}$and at a temperature of localid="1646640967506" $3750\mathrm{K}$.

(b) Calculate the total energy output of the bulb in terms of localid="1646640979854" $\mathrm{W}/{\mathrm{cm}}^2$.

Short answer

a) $1010\mathrm{nm}$and$773\mathrm{nm}$

b)$3.86\times {10}^2{\mathrm{Wcm}}^2$and$1.125\times {10}^3\mathrm{W}/{\mathrm{cm}}^2$

Step-by-step solution

 Given information

$a)T=2870\mathrm{K}\mathrm{and}T=3750\mathrm{K}$

Given formula is

$$\begin{gathered} {𝞴}_{\max }\mathrm{T}=2.9\times {10}^3𝞵\mathrm{mK} \\ {\mathrm{E}}_{\mathrm{t}}=5.69\times {10}^{-8}\times {\mathrm{T}}^4 \end{gathered}$$

Part a) Step 1:  Calculations

Putting values in given formula

When $T=2870\mathrm{K}$

$$\begin{gathered} {𝞴}_{\max }\mathrm{T}=2.9\times {10}^3𝞵\mathrm{mK} \\ {𝞴}_{\max }\times 2870=2.9\times {10}^3 \\ {𝞴}_{\max }=\frac{2.9\times {10}^3}{2870} \\ {𝞴}_{\max }=1.010𝞵\mathrm{m} \\ {𝞴}_{\max }=1010\times {10}^{-3}\times {10}^{-6}\mathrm{m} \\ {𝞴}_{\max }=1010\times {10}^{-9}\mathrm{m} \\ {𝞴}_{\max }=1010\mathrm{nm} \end{gathered}$$

Part a) Step 2:  Calculations

Putting values in formula

When$\mathrm{T}=3750\mathrm{K}$

$$\begin{gathered} {𝞴}_{\max }\mathrm{T}=2.9\times {10}^3𝞵\mathrm{mK} \\ {𝞴}_{\max }\times 3750=2.9\times {10}^3 \\ {𝞴}_{\max }=\frac{2.9\times {10}^3}{3750} \\ {𝞴}_{\max }=0.7733𝞵\mathrm{m} \\ {𝞴}_{\max }=773.3\times {10}^{-3}\times {10}^{-6}\mathrm{m} \\ {𝞴}_{\max }=773\times {10}^{-9}\mathrm{m} \\ {𝞴}_{\max }=773\mathrm{nm} \end{gathered}$$

Part b) Step 1: Calculations.

By substituting values in formula for total energy

$$\begin{gathered} {\mathrm{E}}_{\mathrm{t}}=5.69\times {10}^{-8}\times {2870}^4 \\ {\mathrm{E}}_{\mathrm{t}}=3.86\times {10}^6\mathrm{W}/{\mathrm{m}}^2 \\ {\mathrm{E}}_{\mathrm{t}}=3.86\times {10}^6\times {10}^{-4}\mathrm{W}/{\mathrm{cm}}^2 \\ {\mathrm{E}}_{\mathrm{t}}=3.86\times {10}^2\mathrm{W}/{\mathrm{cm}}^2 \end{gathered}$$

Part b) Step 2: Calculations.

By substituting values in formula for total energy

$$\begin{gathered} {\mathrm{E}}_{\mathrm{t}}=5.69\times {10}^{-8}\times {3750}^4 \\ {\mathrm{E}}_{\mathrm{t}}=1.125\times {10}^7\mathrm{W}/{\mathrm{m}}^2 \\ {\mathrm{E}}_{\mathrm{t}}=1.125\times {10}^7\times {10}^{-4}\mathrm{W}/{\mathrm{cm}}^2 \\ {\mathrm{E}}_{\mathrm{t}}=1.125\times {10}^3\mathrm{W}/{\mathrm{cm}}^2 \end{gathered}$$