Question

For the thermal radiation from an ideal blackbody radiator with a surface temperature of $\mathbf{2000}\mathbf{\text{\hspace{0.17em}K}}$, let ${\mathit{I}}_{\mathbf{c}}$represent the intensity per unit wavelength according to the classical expression for the spectral radiancy and ${\mathit{I}}_{\mathbf{p}}$represent the corresponding intensity per unit wavelength according to the Planck expression. What is the ratio ${\mathit{I}}_{\mathbf{c}}\mathbf{/}{\mathit{I}}_{\mathbf{p}}$for a wavelength of

(a) $\mathbf{400}\mathbf{\text{\hspace{0.17em}nm}}$ (at the blue end of the visible spectrum) and

(b) $\mathbf{200}\mathbf{\text{\hspace{0.17em}μm}}$(in the far infrared)?

(c) Does the classical expression agree with the Planck expression in the shorter wavelength range or the longer wavelength range?

Short answer

(a) The ratio of $I_c/I_p$for wavelength$400\text{\hspace{0.17em}nm}$ is $3.61\times {10}^6$.

(b) The ratio of$I_c/I_p$ for wavelength $200\text{\hspace{0.17em}μm}$is $1.018$.

(c) The classical expression is agree with the Plank’s expression in the longer wavelength range

Step-by-step solution

Write the given data from the question.

The surface temperature,$T=2000\text{\hspace{0.17em}K}$

The intensity per unit wavelength according to classical radiation is $I_c$.

The intensity per unit wavelength according to plank’s radiation is $I_p$.

Determine the formulas to calculate the ratio Ic/Ip.

The expression for the intensity according to classical radiation is given as follows.

${\mathit{I}}_{\mathbf{c}}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\mathbf{c}\mathbf{k}\mathbf{T}}{{\mathbf{\lambda }}^{\mathbf{4}}}$

Here,cis the speed of light, kis the Boltzmann constant and $\lambda$is the wavelength.

The expression for the intensity according to Plank’s radiation is given as follows.

${\mathit{I}}_{\mathbf{p}}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }{\mathbf{c}}^{\mathbf{2}}\mathbf{h}}{{\mathbf{\lambda }}^{\mathbf{5}}}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{h}\mathbf{c}\mathbf{/}\mathbf{\lambda }\mathbf{k}\mathbf{T}}\mathbf{-}\mathbf{1}}$

Here,h is the plank’s constant.

Calculate the ratio for λ=400 nm .

(a)

The value of Boltzmann constant is,$1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$.

The value of plank’s constant is,$6.62\times {10}^{-34}mkg/s$.

Calculate the ratio of $I_c/I_p$.

$\begin{array}{l}\frac{I_c}{I_p}=\frac{\frac{2\pi ckT}{{\lambda }^4}}{\frac{2\pi c^{2}h}{{\lambda }^5}\frac{1}{e^{hc/\lambda kT}-1}}\\ \frac{I_c}{I_p}=\frac{kT}{\frac{hc}{\lambda }\frac{1}{e^{hc/\lambda kT}-1}}\\ \frac{I_c}{I_p}=\frac{\lambda kT}{hc}(e^{hc/\lambda kT}-1)\end{array}$ …(i)

Substitute$1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$for K , $6.62\times {10}^{-34}mkg/s$for , for h , $3\times {10}^8$for c ,$2000\text{\hspace{0.17em}K}$for T and $400\text{\hspace{0.17em}nm}$for $\lambda$into equation (i).

$\begin{array}{l}\frac{I_c}{I_p}=\frac{400\times {10}^{-9}\text{\hspace{0.17em}m}\times 1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\times 2000\text{\hspace{0.17em}K}}{6.62\times {10}^{-34}mkg/s\times 3\times {10}^8\text{\hspace{0.17em}m}/\text{s}}\left(e^{\frac{6.62\times {10}^{-34}mkg/s\times 3\times {10}^8\text{\hspace{0.17em}m}/\text{s}}{400\times {10}^{-9}\text{\hspace{0.17em}m}\times 1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\times 2000\text{\hspace{0.17em}K}}}-1\right)\\ \frac{I_c}{I_p}=\frac{1104\times {10}^{-29}}{19.86\times {10}^{-26}}\left(e^{\frac{19.86\times {10}^{-26}}{1104\times {10}^{-29}}}-1\right)\\ \frac{I_c}{I_p}=55.58\times {10}^{-3}(e^{17.989}-1)\\ \frac{I_c}{I_p}=55.58\times {10}^{-3}(6.49\times {10}^7-1)\end{array}$

Solve further as,

$\begin{array}{l}\frac{I_c}{I_p}=0.05558\times 6.49\times {10}^7\\ \frac{I_c}{I_p}=0.361\times {10}^7\\ \frac{I_c}{I_p}=3.61\times {10}^6\end{array}$

Hence the ratio of$I_c/I_p$ for wavelength $400\text{\hspace{0.17em}nm}$is $3.61\times {10}^6$.

Calculate the ratio Ic/Ip for  λ=200 μm.

(b)

Calculate the ratio of $I_c/I_p$.

Substitute$1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}$for K, $6.62\times {10}^{-34}mkg/s$for h , $3\times {10}^8$for c, $2000\text{\hspace{0.17em}K}$for T and $200\text{\hspace{0.17em}μm}$for $\lambda$into equation (i).

$\begin{array}{l}\frac{I_c}{I_p}=\frac{200\times {10}^{-6}\text{\hspace{0.17em}m}\times 1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\times 2000\text{\hspace{0.17em}K}}{6.62\times {10}^{-34}mkg/s\times 3\times {10}^8\text{\hspace{0.17em}m}/\text{s}}\left(e^{\frac{6.62\times {10}^{-34}mkg/s\times 3\times {10}^8\text{\hspace{0.17em}m}/\text{s}}{200\times {10}^{-6}\text{\hspace{0.17em}m}\times 1.38\times {10}^{-23}\text{\hspace{0.17em}J}/\text{K}\times 2000\text{\hspace{0.17em}K}}}-1\right)\\ \frac{I_c}{I_p}=\frac{552\times {10}^{-26}}{19.86\times {10}^{-26}}\left(e^{\frac{19.86\times {10}^{-26}}{552\times {10}^{-26}}}-1\right)\\ \frac{I_c}{I_p}=27.79(e^{0.0359}-1)\\ \frac{I_c}{I_p}=27.79(1.0366-1)\end{array}$

Solve further as,

$\begin{array}{l}\frac{I_c}{I_p}=27.79\times 0.0366\\ \frac{I_c}{I_p}=1.018\end{array}$

Hence the ratio of $I_c/I_p$for wavelength $200\mu \text{m}$is$1.018$

 Step 5: Determine that classical expression agree with Plank’s expression in the longer or shorter wavelength range  

(c)

From the result of the part (b), it is clear that the classical expression is agree with the Plank’s expression in the longer wavelength range.

Hence the agreement between $I_c$and $I_p$is at longer wavelength range.