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 .

Question

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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