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?

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 $I_p$.

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.

(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 $3.61\times {10}^6$.

(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$1.018$

(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.