The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

Question

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Answer 1

Question

Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

(a)

Expert Solution

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

l1(λ,t)l2(λ,t)=2πhc2λ151[ehc/λ1kT−1]2πhc2λ251[ehc/λ2kT−1] (I)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Conclusion:

Substitute 400 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(4×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.073

Substitute 700 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(7×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.754

Thus, the fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

(b)

To determine

The ratio of intensity at 400 nm to 700 nm wavelength with maximum intensity.

(b)

Expert Solution

Explanation of Solution

Write the expression for the Planck’s law of radiation.

R=2πc2h∫λ1λ2e(−hc/λkT)λ−5dλ (II)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

P=σT4 (III)

Here, P is the power emitted and σ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

r=RP (IV)

Here, r is the fraction of the power.

Conclusion:

Substitute 400 nm for λ1, 700 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (II).

R=2π(3×108 m/s)2(6.6×10−34 J⋅s)∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.73×10−16∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.71×105 W/m2

Substitute 5.66×10−8 W/m2K4 for σ and 3000 K for T in equation (III).

P=5.66×10−8 W/m2K4(3000 K)4=4.59×106 W/m2

Substitute 4.59×106 W/m2 for P and 3.71×105 W/m2 for R in equation (IV).

r=3.71×105 W/m24.59×106 W/m2=0.081

Thus, the fraction of the visible power to the maximum intensity power is 0.081.

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Answer 2

Question

Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

(a)

Expert Solution

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

l1(λ,t)l2(λ,t)=2πhc2λ151[ehc/λ1kT−1]2πhc2λ251[ehc/λ2kT−1] (I)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Conclusion:

Substitute 400 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(4×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.073

Substitute 700 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(7×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.754

Thus, the fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

(b)

To determine

The ratio of intensity at 400 nm to 700 nm wavelength with maximum intensity.

(b)

Expert Solution

Explanation of Solution

Write the expression for the Planck’s law of radiation.

R=2πc2h∫λ1λ2e(−hc/λkT)λ−5dλ (II)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

P=σT4 (III)

Here, P is the power emitted and σ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

r=RP (IV)

Here, r is the fraction of the power.

Conclusion:

Substitute 400 nm for λ1, 700 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (II).

R=2π(3×108 m/s)2(6.6×10−34 J⋅s)∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.73×10−16∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.71×105 W/m2

Substitute 5.66×10−8 W/m2K4 for σ and 3000 K for T in equation (III).

P=5.66×10−8 W/m2K4(3000 K)4=4.59×106 W/m2

Substitute 4.59×106 W/m2 for P and 3.71×105 W/m2 for R in equation (IV).

r=3.71×105 W/m24.59×106 W/m2=0.081

Thus, the fraction of the visible power to the maximum intensity power is 0.081.

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Answer 3

Question

Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

(a)

Expert Solution

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

l1(λ,t)l2(λ,t)=2πhc2λ151[ehc/λ1kT−1]2πhc2λ251[ehc/λ2kT−1] (I)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Conclusion:

Substitute 400 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(4×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.073

Substitute 700 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(7×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.754

Thus, the fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

(b)

To determine

The ratio of intensity at 400 nm to 700 nm wavelength with maximum intensity.

(b)

Expert Solution

Explanation of Solution

Write the expression for the Planck’s law of radiation.

R=2πc2h∫λ1λ2e(−hc/λkT)λ−5dλ (II)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

P=σT4 (III)

Here, P is the power emitted and σ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

r=RP (IV)

Here, r is the fraction of the power.

Conclusion:

Substitute 400 nm for λ1, 700 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (II).

R=2π(3×108 m/s)2(6.6×10−34 J⋅s)∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.73×10−16∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.71×105 W/m2

Substitute 5.66×10−8 W/m2K4 for σ and 3000 K for T in equation (III).

P=5.66×10−8 W/m2K4(3000 K)4=4.59×106 W/m2

Substitute 4.59×106 W/m2 for P and 3.71×105 W/m2 for R in equation (IV).

r=3.71×105 W/m24.59×106 W/m2=0.081

Thus, the fraction of the visible power to the maximum intensity power is 0.081.

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Answer 4

Question

Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

(a)

Expert Solution

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

l1(λ,t)l2(λ,t)=2πhc2λ151[ehc/λ1kT−1]2πhc2λ251[ehc/λ2kT−1] (I)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Conclusion:

Substitute 400 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(4×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.073

Substitute 700 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(7×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.754

Thus, the fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

(b)

To determine

The ratio of intensity at 400 nm to 700 nm wavelength with maximum intensity.

(b)

Expert Solution

Explanation of Solution

Write the expression for the Planck’s law of radiation.

R=2πc2h∫λ1λ2e(−hc/λkT)λ−5dλ (II)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

P=σT4 (III)

Here, P is the power emitted and σ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

r=RP (IV)

Here, r is the fraction of the power.

Conclusion:

Substitute 400 nm for λ1, 700 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (II).

R=2π(3×108 m/s)2(6.6×10−34 J⋅s)∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.73×10−16∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.71×105 W/m2

Substitute 5.66×10−8 W/m2K4 for σ and 3000 K for T in equation (III).

P=5.66×10−8 W/m2K4(3000 K)4=4.59×106 W/m2

Substitute 4.59×106 W/m2 for P and 3.71×105 W/m2 for R in equation (IV).

r=3.71×105 W/m24.59×106 W/m2=0.081

Thus, the fraction of the visible power to the maximum intensity power is 0.081.

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Answer 5

Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

(a)

Expert Solution

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

l1(λ,t)l2(λ,t)=2πhc2λ151[ehc/λ1kT−1]2πhc2λ251[ehc/λ2kT−1] (I)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Conclusion:

Substitute 400 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(4×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.073

Substitute 700 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(7×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.754

Thus, the fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

(b)

To determine

The ratio of intensity at 400 nm to 700 nm wavelength with maximum intensity.

(b)

Expert Solution

Explanation of Solution

Write the expression for the Planck’s law of radiation.

R=2πc2h∫λ1λ2e(−hc/λkT)λ−5dλ (II)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

P=σT4 (III)

Here, P is the power emitted and σ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

r=RP (IV)

Here, r is the fraction of the power.

Conclusion:

Substitute 400 nm for λ1, 700 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (II).

R=2π(3×108 m/s)2(6.6×10−34 J⋅s)∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.73×10−16∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.71×105 W/m2

Substitute 5.66×10−8 W/m2K4 for σ and 3000 K for T in equation (III).

P=5.66×10−8 W/m2K4(3000 K)4=4.59×106 W/m2

Substitute 4.59×106 W/m2 for P and 3.71×105 W/m2 for R in equation (IV).

r=3.71×105 W/m24.59×106 W/m2=0.081

Thus, the fraction of the visible power to the maximum intensity power is 0.081.

Answer 6

Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

(a)

Expert Solution

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

l1(λ,t)l2(λ,t)=2πhc2λ151[ehc/λ1kT−1]2πhc2λ251[ehc/λ2kT−1] (I)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Conclusion:

Substitute 400 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(4×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.073

Substitute 700 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(7×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.754

Thus, the fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Answer 7

Chapter 3, Problem 24P

(a)

To determine

The power of the emission in the visible region of the spectrum and compare it with ultraviolet and infrared region.

Answer 8

(a)

Expert Solution

Answer to Problem 24P

The fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the ratio of intensities from Planck’s law for the different wavelength.

l1(λ,t)l2(λ,t)=2πhc2λ151[ehc/λ1kT−1]2πhc2λ251[ehc/λ2kT−1] (I)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Conclusion:

Substitute 400 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(4×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.073

Substitute 700 nm for λ1, 966 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (I).

l1(λ,t)l2(λ,t)=2π(3×108 m/s)2(6.6×10−34 J⋅s)(4×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(7×10−7)(1.38×10−23)3000−1]2π(3×108 m/s)2(6.6×10−34 J⋅s)(9.66×10−7)51[e(3×108 m/s)(6.6×10−34 J⋅s)/(9.66×10−7)(1.38×10−23)3000−1]=0.754

Thus, the fraction of the visible power to the ultraviolet and infrared power is 0.073 and 0.754 respectively.

Answer 13

(b)

To determine

The ratio of intensity at 400 nm to 700 nm wavelength with maximum intensity.

(b)

Expert Solution

Explanation of Solution

Write the expression for the Planck’s law of radiation.

R=2πc2h∫λ1λ2e(−hc/λkT)λ−5dλ (II)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

P=σT4 (III)

Here, P is the power emitted and σ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

r=RP (IV)

Here, r is the fraction of the power.

Conclusion:

Substitute 400 nm for λ1, 700 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (II).

R=2π(3×108 m/s)2(6.6×10−34 J⋅s)∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.73×10−16∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.71×105 W/m2

Substitute 5.66×10−8 W/m2K4 for σ and 3000 K for T in equation (III).

P=5.66×10−8 W/m2K4(3000 K)4=4.59×106 W/m2

Substitute 4.59×106 W/m2 for P and 3.71×105 W/m2 for R in equation (IV).

r=3.71×105 W/m24.59×106 W/m2=0.081

Thus, the fraction of the visible power to the maximum intensity power is 0.081.

Answer 14

(b)

To determine

The ratio of intensity at 400 nm to 700 nm wavelength with maximum intensity.

Answer 15

(b)

Expert Solution

Explanation of Solution

Write the expression for the Planck’s law of radiation.

R=2πc2h∫λ1λ2e(−hc/λkT)λ−5dλ (II)

Here, c is the speed of light, h is the Planck’s constant, k is the Boltzmann’s constant and λ is the wavelength of emission.

Write the expression for the total power radiated by the lamp.

P=σT4 (III)

Here, P is the power emitted and σ is the Stefan’s constant.

Write the expression for the fraction of power in the visible region.

r=RP (IV)

Here, r is the fraction of the power.

Conclusion:

Substitute 400 nm for λ1, 700 nm for λ2, 1.38×10−23 J/K for k, 1.986×10−25 J⋅m for hc and 3000 K for T in equation (II).

R=2π(3×108 m/s)2(6.6×10−34 J⋅s)∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.73×10−16∫4×10−77×10−7e(−1.986×10−25 J⋅m/λ(4.14×10−20))λ−5dλ=3.71×105 W/m2