Calculate the power of light in the wavelength range λ = 350 nm − 351 nm (that is, let d λ be Δ λ = 1 nm in the Planck’s law, and let λ be 350.5 nm ) at temperatures of 1000 K , 3000 K and 10 , 000 K .

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Answer 1

Textbook Question

Chapter 9, Problem 9.32E

Calculate the power of light in the wavelength range $λ = 350 nm − 351 nm$ (that is, let $d λ$ be $Δ λ = 1 nm$ in the Planck’s law, and let $λ$ be $350.5 nm$ ) at temperatures of $1000 K$ , $3000 K$ and $10 , 000 K$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $λ=350−351 nm$ at temperatures $1000 K$ , $3000 K$ and $10,000 K$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Answer to Problem 9.32E

The power of light in the wavelength range $λ=350−351 nm$ at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Explanation of Solution

It is given that wavelength range is $350 nm−351 nm$ , the value of $dλ$ is $1 nm$ and $λ$ is $350.5 nm$ .

The formula to calculate the power of light is,

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Where,

• $ξ$ is the power flux.

• $λ$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108m s−1)2(350.5 ×10−9 m)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×1000 K−1)×(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−27−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×17.00×1017−1×1×10−9 J⋅s−1m−2=7.08×1016×1.428×10−18×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.01×10−10 W/m2$ .

Substitute the values of constants and temperature $3000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×3000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−251.451×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×18.8×105−1×1×10−9 J⋅s−1m−2=7.08×1016×1.133×10−6×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $80.21 W/m2$ .

Substitute the values of constants and temperature $10,000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×10000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×160.88−1×1×10−9 J⋅s−1m−2=7.08×1016×0.0167×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.18×106 W/m2$ .

Conclusion

The power of light at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

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Answer 2

Textbook Question

Chapter 9, Problem 9.32E

Calculate the power of light in the wavelength range $λ = 350 nm − 351 nm$ (that is, let $d λ$ be $Δ λ = 1 nm$ in the Planck’s law, and let $λ$ be $350.5 nm$ ) at temperatures of $1000 K$ , $3000 K$ and $10 , 000 K$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $λ=350−351 nm$ at temperatures $1000 K$ , $3000 K$ and $10,000 K$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Answer to Problem 9.32E

The power of light in the wavelength range $λ=350−351 nm$ at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Explanation of Solution

It is given that wavelength range is $350 nm−351 nm$ , the value of $dλ$ is $1 nm$ and $λ$ is $350.5 nm$ .

The formula to calculate the power of light is,

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Where,

• $ξ$ is the power flux.

• $λ$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108m s−1)2(350.5 ×10−9 m)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×1000 K−1)×(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−27−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×17.00×1017−1×1×10−9 J⋅s−1m−2=7.08×1016×1.428×10−18×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.01×10−10 W/m2$ .

Substitute the values of constants and temperature $3000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×3000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−251.451×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×18.8×105−1×1×10−9 J⋅s−1m−2=7.08×1016×1.133×10−6×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $80.21 W/m2$ .

Substitute the values of constants and temperature $10,000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×10000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×160.88−1×1×10−9 J⋅s−1m−2=7.08×1016×0.0167×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.18×106 W/m2$ .

Conclusion

The power of light at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

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Answer 3

Textbook Question

Chapter 9, Problem 9.32E

Calculate the power of light in the wavelength range $λ = 350 nm − 351 nm$ (that is, let $d λ$ be $Δ λ = 1 nm$ in the Planck’s law, and let $λ$ be $350.5 nm$ ) at temperatures of $1000 K$ , $3000 K$ and $10 , 000 K$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $λ=350−351 nm$ at temperatures $1000 K$ , $3000 K$ and $10,000 K$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Answer to Problem 9.32E

The power of light in the wavelength range $λ=350−351 nm$ at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Explanation of Solution

It is given that wavelength range is $350 nm−351 nm$ , the value of $dλ$ is $1 nm$ and $λ$ is $350.5 nm$ .

The formula to calculate the power of light is,

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Where,

• $ξ$ is the power flux.

• $λ$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108m s−1)2(350.5 ×10−9 m)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×1000 K−1)×(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−27−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×17.00×1017−1×1×10−9 J⋅s−1m−2=7.08×1016×1.428×10−18×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.01×10−10 W/m2$ .

Substitute the values of constants and temperature $3000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×3000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−251.451×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×18.8×105−1×1×10−9 J⋅s−1m−2=7.08×1016×1.133×10−6×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $80.21 W/m2$ .

Substitute the values of constants and temperature $10,000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×10000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×160.88−1×1×10−9 J⋅s−1m−2=7.08×1016×0.0167×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.18×106 W/m2$ .

Conclusion

The power of light at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

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Answer 4

Textbook Question

Chapter 9, Problem 9.32E

Calculate the power of light in the wavelength range $λ = 350 nm − 351 nm$ (that is, let $d λ$ be $Δ λ = 1 nm$ in the Planck’s law, and let $λ$ be $350.5 nm$ ) at temperatures of $1000 K$ , $3000 K$ and $10 , 000 K$ .

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $λ=350−351 nm$ at temperatures $1000 K$ , $3000 K$ and $10,000 K$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Answer to Problem 9.32E

The power of light in the wavelength range $λ=350−351 nm$ at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Explanation of Solution

It is given that wavelength range is $350 nm−351 nm$ , the value of $dλ$ is $1 nm$ and $λ$ is $350.5 nm$ .

The formula to calculate the power of light is,

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Where,

• $ξ$ is the power flux.

• $λ$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108m s−1)2(350.5 ×10−9 m)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×1000 K−1)×(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−27−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×17.00×1017−1×1×10−9 J⋅s−1m−2=7.08×1016×1.428×10−18×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.01×10−10 W/m2$ .

Substitute the values of constants and temperature $3000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×3000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−251.451×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×18.8×105−1×1×10−9 J⋅s−1m−2=7.08×1016×1.133×10−6×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $80.21 W/m2$ .

Substitute the values of constants and temperature $10,000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×10000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×160.88−1×1×10−9 J⋅s−1m−2=7.08×1016×0.0167×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.18×106 W/m2$ .

Conclusion

The power of light at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $λ=350−351 nm$ at temperatures $1000 K$ , $3000 K$ and $10,000 K$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Answer to Problem 9.32E

The power of light in the wavelength range $λ=350−351 nm$ at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Explanation of Solution

It is given that wavelength range is $350 nm−351 nm$ , the value of $dλ$ is $1 nm$ and $λ$ is $350.5 nm$ .

The formula to calculate the power of light is,

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Where,

• $ξ$ is the power flux.

• $λ$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108m s−1)2(350.5 ×10−9 m)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×1000 K−1)×(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−27−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×17.00×1017−1×1×10−9 J⋅s−1m−2=7.08×1016×1.428×10−18×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.01×10−10 W/m2$ .

Substitute the values of constants and temperature $3000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×3000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−251.451×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×18.8×105−1×1×10−9 J⋅s−1m−2=7.08×1016×1.133×10−6×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $80.21 W/m2$ .

Substitute the values of constants and temperature $10,000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×10000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×160.88−1×1×10−9 J⋅s−1m−2=7.08×1016×0.0167×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.18×106 W/m2$ .

Conclusion

The power of light at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $λ=350−351 nm$ at temperatures $1000 K$ , $3000 K$ and $10,000 K$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Answer to Problem 9.32E

The power of light in the wavelength range $λ=350−351 nm$ at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Explanation of Solution

It is given that wavelength range is $350 nm−351 nm$ , the value of $dλ$ is $1 nm$ and $λ$ is $350.5 nm$ .

The formula to calculate the power of light is,

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Where,

• $ξ$ is the power flux.

• $λ$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108m s−1)2(350.5 ×10−9 m)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×1000 K−1)×(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−27−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×17.00×1017−1×1×10−9 J⋅s−1m−2=7.08×1016×1.428×10−18×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.01×10−10 W/m2$ .

Substitute the values of constants and temperature $3000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×3000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−251.451×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×18.8×105−1×1×10−9 J⋅s−1m−2=7.08×1016×1.133×10−6×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $80.21 W/m2$ .

Substitute the values of constants and temperature $10,000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×10000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×160.88−1×1×10−9 J⋅s−1m−2=7.08×1016×0.0167×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.18×106 W/m2$ .

Conclusion

The power of light at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The power of light in the wavelength range $λ=350−351 nm$ at temperatures $1000 K$ , $3000 K$ and $10,000 K$ is to be calculated.

Concept introduction:

Planck’s equation can also be represented in the form of power flux or infinitesimal power per unit area for a black body radiation at a given temperature and wavelengths as shown below.

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Answer 12

Answer to Problem 9.32E

The power of light in the wavelength range $λ=350−351 nm$ at the temperatures $1000 K$ , $3000 K$ and $10,000 K$ is $1.01×10−10 W/m2$ , $80.21 W/m2$ and $1.18×106 W/m2$ respectively.

Answer 13

Explanation of Solution

It is given that wavelength range is $350 nm−351 nm$ , the value of $dλ$ is $1 nm$ and $λ$ is $350.5 nm$ .

The formula to calculate the power of light is,

$dξ=2πhc2λ5(1ehc/λkT−1)dλ$

Where,

• $ξ$ is the power flux.

• $λ$ is the wavelength.

• $h$ is the Planck’s constant.

• $k$ is the Boltzmann constant.

• $c$ is the speed of light.

Substitute the values of constants and temperature $1000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108m s−1)2(350.5 ×10−9 m)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×1000 K−1)×(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−27−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×17.00×1017−1×1×10−9 J⋅s−1m−2=7.08×1016×1.428×10−18×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $1.01×10−10 W/m2$ .

Substitute the values of constants and temperature $3000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×3000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−251.451×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×18.8×105−1×1×10−9 J⋅s−1m−2=7.08×1016×1.133×10−6×1×10−9 J⋅s−1m−2$

On simplifying the above equation,

$1 W=1 J⋅s−1$

$dξ=1.01×10−10 W/m2$

Thus, the value of power flux is $80.21 W/m2$ .

Substitute the values of constants and temperature $10,000 K$ to calculate the power of light in the given formula.

$dξ=(2(3.14)(6.626×10−34 J⋅s)(3×108ms−1)2(350.5 ×10−9)5(1e6.626×10−34 J⋅s×3×108ms−1350.5 ×10−9m×1.38×10−23 JK-1×10000 K−1)(1×10−9m))=3.745×10−165.289×10−33×(1e1.9878×10−254.8369×10−26−1)×1×10−9 J⋅s−1m−2=3.745×10−165.289×10−33×160.88−1×1×10−9 J⋅s−1m−2=7.08×1016×0.0167×1×10−9 J⋅s−1m−2$