91影视

Relation between wavelength and frequency

$\mathbf{v}\mathbf{=}\mathbf{位蹿}\mathbf{\text{鈥�}}\mathbf{鈥�}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$

Explanation of Solution

Given Information:

$\begin{array}{l}\frac{\mathrm{dU}}{\mathrm{df}}=\frac{{\mathrm{hf}}^3}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳\frac{8\mathrm{蟺痴}}{{\mathrm{c}}^3}{\mathrm{f}}^2\\ {\mathrm{位}}_{\max }\mathrm{T}=2.898脳{10}^{鈭�3}{\text{m}}^2\mathrm{K}\end{array}$

Formula Used:

We know,

$\mathrm{v}=\mathrm{位蹿}\text{鈥�}鈥�..\left(2\right)$

First, it would help to rewrite equation (1) in terms of wavelength instead of frequency. It can be simplified a little with using equation (2) to fill in forand then simplify:

$\begin{array}{l}\mathrm{dU}=\frac{{\mathrm{hf}}^3}{\frac{\mathrm{hf}}{{\mathrm{k}}^{\mathrm{T}}\mathrm{T}}}\frac{8\mathrm{蟺痴}}{{\mathrm{c}}^3}\mathrm{df}\\ =\frac{{\mathrm{hf}}^3}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}\frac{8\mathrm{蟺痴}}{(\mathrm{位蹿}}\mathrm{df}}\\ =\frac{{\mathrm{hf}}^3}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}\frac{8\mathrm{蟺痴}}{{\mathrm{位}}^3{\mathrm{f}}^3}\mathrm{df}\\ \mathrm{dU}=\frac{\mathrm{h}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}\frac{8\mathrm{蟺痴}}{{\mathrm{位}}^3}\mathrm{df}\end{array}$

Then use equation (2) to get an expression for the frequencyand the infinitesimal unit of frequency

$\begin{array}{l}\mathrm{c}=\mathrm{位蹿}\\ \mathrm{f}=\frac{\mathrm{c}}{\mathrm{位}}\\ \mathrm{df}=鈭�\frac{\mathrm{c}}{{\mathrm{位}}^2}\mathrm{诲位}\end{array}$

Though since we're just interested in the magnitude, we can just simplify that as:

$\mathrm{df}=鈭�\frac{\mathrm{c}}{{\mathrm{位}}^2}\mathrm{诲位}$

Those can then be inserted in forandin equation (3)

$\begin{array}{l}\mathrm{dU}=\frac{\mathrm{h}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}\frac{8\mathrm{蟺痴}}{{\mathrm{位}}^3}\mathrm{df}\\ =\frac{\mathrm{hc}}{{\mathrm{e}}^{\frac{\mathrm{h}}{{\mathrm{位办}}_{\mathrm{B}}\mathrm{T}}\left(\frac{\mathrm{c}}{\mathrm{位}}\right)}\frac{8\mathrm{蟺痴}}{{\mathrm{位}}^3}}\mathrm{诲位}\end{array}$

Rewrite it slightly as given below:

$\frac{\mathrm{dU}}{\mathrm{诲位}}=\frac{\mathrm{dU}}{\mathrm{诲位}}\frac{\mathrm{df}}{\mathrm{诲位}}$

$=\left(\frac{\mathrm{hf}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳\frac{8\mathrm{蟺痴}}{{\mathrm{c}}^3}{\mathrm{f}}^2\right)\left(鈭�\frac{\mathrm{c}}{{\mathrm{位}}^2}\right)$

Though since we're only interested in the magnitude, it can be written as:

$\frac{\mathrm{dU}}{\mathrm{诲位}}=\left(\frac{\mathrm{hf}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳\frac{8\mathrm{蟺痴}}{{\mathrm{c}}^3}{\mathrm{f}}^2\right)\left(\frac{\mathrm{c}}{{\mathrm{位}}^2}\right)$

And then simplifying that yields:

$\begin{array}{l}\frac{\mathrm{dU}}{\mathrm{诲位}}=\left(\frac{\mathrm{hf}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳\frac{8\mathrm{蟺痴}}{{\mathrm{c}}^3}{\mathrm{f}}^2\right)\left(\frac{\mathrm{c}}{{\mathrm{位}}^2}\right)\\ =\frac{\mathrm{hf}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳8\mathrm{蟺痴}脳{\left(\frac{\mathrm{f}}{\mathrm{c}}\right)}^3脳\frac{1}{{\mathrm{位}}^2}\\ =\frac{8\mathrm{蟺痴hc}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳\left(\frac{1}{{\mathrm{位}}^3}\right)脳\frac{1}{{\mathrm{位}}^2}\\ =\frac{8\mathrm{蟺痴hc}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳\frac{1}{{\mathrm{位}}^5}\end{array}$

So Planck's spectral energy density in terms of wavelength is

$\frac{\mathrm{dU}}{\mathrm{诲位}}=\frac{8\mathrm{蟺痴hc}}{{\mathrm{e}}^{\frac{\mathrm{hf}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}}鈭�1}脳\frac{1}{{\mathrm{位}}^5}$

Since the maximum wavelength is needed, the derivative of that with respect too wavelength needs to be taken. But since we've rewritten it, we'll use the chain rule in order to do so:

$\frac{鈭�}{鈭�位}\left(\frac{dU}{d位}\right)=\left[\frac{鈭�}{鈭�位}\left(\frac{dU}{d位}\right)\right]\left[\frac{鈭�x}{鈭�位}\right]$

So for taking the partial derivative with respect to for equation (6), we get:

$\frac{鈭�}{鈭�\mathrm{位}}\left(\frac{\mathrm{dU}}{\mathrm{诲位}}\right)=\left[\frac{鈭�8{\mathrm{蟺痴hce}}^{\mathrm{x}}}{\left({\mathrm{e}}^{\mathrm{x}}鈭�1\right)}\right]\left[\left(\frac{{\mathrm{虫魏}}_{\mathrm{B}}\mathrm{T}}{\mathrm{鈩�}\mathrm{c}}\right)\right]+\left[\frac{8\mathrm{蟺痴hce}}{\left({\mathrm{e}}^{\mathrm{x}}鈭�1\right)}\right]\left[5{\mathrm{x}}^4{\left(\frac{{\mathrm{魏}}_{\mathrm{B}}\mathrm{T}}{\mathrm{hc}}\right)}^5\right]$

Since we want to find where:

$\frac{鈭�}{鈭�\mathrm{位}}\left(\frac{\mathrm{dU}}{\mathrm{诲位}}\right)=0$

That can be found where

$\frac{\frac{鈭�}{鈭�位}\left(\frac{dU}{d位}\right)=0}{}$

Since multiplying by the partial of with respect to will still give 0 . So insert 0 for the left side of equation (7) after simplifying it a little:

$\begin{array}{l}\frac{鈭�}{鈭�位}\left(\frac{dU}{d位}\right)=\left[\frac{鈭�8蟺Vhc{e}^x}{\left(e^x鈭�1\right)}\right]\left[\left(\frac{x{魏}_{B}T}{hc}\right)\right]+\left[\frac{8蟺Vhce}{\left(e^x鈭�1\right)}\right]\left[5x^4{\left(\frac{{魏}_{B}T}{hc}\right)}^5\right]\\ \frac{鈭�}{鈭�位}\left(\frac{dU}{d位}\right)=5x^4{\left(\frac{{魏}_{B}T}{hc}\right)}^5\left[\frac{8蟺Vhc}{\left(e^x鈭�1\right)}\right]鈭�{\left[\left(\frac{x{魏}_{B}T}{hc}\right)\right]}^5\left[\frac{8蟺Vhc}{{\left(e^x鈭�1\right)}^5}\right]0\\ =5x^4{\left(\frac{{魏}_{B}T}{hc}\right)}^5\left[\frac{8蟺Vhc}{\left(e^x鈭�1\right)}\right]鈭�{\left[\left(\frac{x{魏}_{B}T}{hc}\right)\right]}^5\left[\frac{8蟺Vh{c}^x}{{\left(e^x鈭�1\right)}^2}\right]0\\ =x^4\left(\frac{8蟺Vhc}{\left(e^x鈭�1\right)}\right){\left(\frac{{魏}_{B}T}{hc}\right)}^5\left[5鈭�\frac{x{e}^x}{e^x鈭�1}\right]\end{array}$

Since the expression in front of the bracket is just some constant, it can drop out, leaving:

$\begin{array}{l}0={\mathrm{x}}^4\left(\frac{8\mathrm{蟺痴hc}}{\left({\mathrm{e}}^{\mathrm{x}}鈭�1\right)}\right){\left(\frac{{\mathrm{魏}}_{\mathrm{B}}\mathrm{T}}{\mathrm{hc}}\right)}^5\left[5鈭�\frac{{\mathrm{xe}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}鈭�1}\right]\\ 0=5鈭�\frac{{\mathrm{xe}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}鈭�1}\\ 5=\frac{{\mathrm{xe}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}鈭�1}\end{array}$

Since that's a transcendental equation, it needs to be solved numerically. Doing so gives:

$x=4.965$

So now that is known, equation (5) can be rewritten:

$\begin{array}{l}\mathrm{x}=\frac{\mathrm{hc}}{{\mathrm{位魏}}_{\mathrm{B}}\mathrm{T}}\\ \mathrm{位罢}=\frac{\mathrm{hc}}{{\mathrm{位魏}}_{\mathrm{R}}}\end{array}$W

Substitute $\left(1.38脳{10}^{鈭�23}{\text{m}}^2脳{\text{kg}}_{\text{s}}{\text{s}}^{鈭�2}\right)$ for ${\mathrm{魏}}_{\mathrm{B}}$is Boltzmann's constant,

$\left(6.63脳{10}^{鈭�34}\right.$J.s ) for Planck's constant $\mathrm{h}$, and $\left(3脳{10}^8\text{m}\right)$for the speed of light in vacuum c in the equation we get.$\begin{array}{l}\mathrm{位罢}=\frac{\mathrm{hc}}{{\mathrm{位魏}}_{\mathrm{B}}}\\ =\frac{\left(6.63脳{10}^{鈭�34}\text{J}.\text{s}\right)\left(3.0脳{10}^8\frac{\text{m}}{\text{s}}\right)}{(4.965)\left(1.38脳{10}^{鈭�23}\frac{\text{J}}{\text{K}}\right)}\\ =2.90脳{10}^{鈭�3}\text{m}.\text{K}\end{array}$