Solution to Set 1

Problem 1

Part a

It is a common misconception that Max Planck derived his now-famous law, Planck's Law, in order to resolve the so-called "ultraviolet catastrophe," which predicts, from classical physics, that a blackbody will emit greater and greater intensity radiation at shorter wavelengths, thus outputting infinite power as electromagnetic radiation. In fact, this problem was not noticed until five years after Planck derived his law.

What actually motivated Planck was his desire to improve on the Wien approximation, which fit known blackbody radiation spectra only at short wavelengths. Expressed as spectral radiance:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(\lambda, T) = \frac{2 h c^2} {\lambda^5} e^{-\frac{hc}{\lambda kT}}}

Conversely, the Rayleigh-Jeans law fit the data only at long wavelengths:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(\lambda, T) = \frac{2 c k T}{\lambda^4}}

Planck derived his function to fit the data at all wavelengths:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(\lambda,T) =\frac{2 hc^2}{\lambda^5}\frac{1}{ e^{\frac{hc}{\lambda kT}}-1}}

Planck derived his law via a consideration of various ways in which electromagnetic energy can be distributed over the different modes of oscillation of charged oscillators in matter (today known to be atoms). He found that when he assumed the energy to be quantized, the above law emerged and fit the data very well.

Part b

Planck's law, expressed as spectral radiance in terms of wavelength and temperature:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(\lambda,T) =\frac{2hc^2}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT}}-1}}

From the above, we will derive Wien's displacement law:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_{\mathrm{max}} = \frac{b}{T}}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_{\mathrm{max}}} is the wavelength of maximum intensity electromagnetic radiation output for a blackbody in thermal equilibrium at absolute temperature Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} is a proportionality constant (for our purposes here, it will not be necessary to calculate the exact value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} , only to show that it must be a constant).

Differentiating Planck's law with respect to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dI}{d\lambda} = 2hc^2\left(-\frac{5}{\lambda^6}\frac{1}{e^{\frac{hc}{\lambda kT}}-1} + \frac{1}{\lambda^5}\frac{hc}{kT}\frac{1}{\lambda^2}\frac{e^{\frac{hc}{\lambda kT}}}{(e^{\frac{hc}{\lambda kT}}-1)^2}\right) = 2hc^2\left(-\frac{5}{\lambda^6}\frac{1}{e^{\frac{hc}{\lambda kT}}-1} + \frac{hc}{\lambda^7 kT}\frac{e^{\frac{hc}{\lambda kT}}}{(e^{\frac{hc}{\lambda kT}}-1)^2}\right)}

To find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_{\mathrm{max}}} , we set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{dI}{d\lambda} = 0} and solve:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2hc^2\left(-\frac{5}{\lambda_{\mathrm{max}}^6}\frac{1}{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}-1} + \frac{hc}{\lambda_{\mathrm{max}}^7 kT}\frac{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}}{(e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}-1)^2}\right) = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{5}{\lambda_{\mathrm{max}}^6}\frac{1}{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}-1} + \frac{hc}{\lambda_{\mathrm{max}}^7 kT}\frac{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}}{(e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}-1)^2} = 0}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{hc}{\lambda_{\mathrm{max}}^7 kT}\frac{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}}{(e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}-1)^2} = \frac{5}{\lambda_{\mathrm{max}}^6}\frac{1}{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}-1}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{hc}{\lambda_{\mathrm{max}} kT}\frac{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}}{e^{\frac{hc}{\lambda_{\mathrm{max}} kT}}-1} = 5}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{hc}{\lambda_{\mathrm{max}} kT}\frac{1}{1-e^{-\frac{hc}{\lambda_{\mathrm{max}} kT}}} = 5}

Now, for simplicity we define:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi \equiv \frac{hc}{\lambda_{\mathrm{max}} kT}}

and we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\chi}{1-e^{-\chi}} = 5}

It is clear from the above equation that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi} must be a constant. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi} is a constant, then examination of its definition above reveals only two non-constant terms: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T} . We thus rearrange the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi} equation to give a simple relation between these two variables:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_{\mathrm{max}} = \frac{hc}{k\chi}\frac{1}{T}}

Defining a constant of proportionality Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b \equiv \frac{hc}{k\chi}}

We now have Wien's displacement law in its most general form:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_{\mathrm{max}} = \frac{b}{T}}

Part c

The Stefan-Boltzmann law states that the total irradiance (total energy radiated per unit surface area per unit time), j^*, emitted by a blackbody in thermal equilibrium at temperature Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T} is given as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\star = \sigma T^4}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma} is a proportionality constant. We will now derive this relation from Planck's law, expressed as spectral radiance in terms of radiant frequency and temperature:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(\nu,T) = \frac{2 h\nu^3}{c^2}\frac{1}{ e^{\frac{h\nu}{kT}}-1}}

In order to obtain the total irradiance of a blackbody following Planck's law, we must integrate its spectral radiance Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} over a half-sphere of solid angle and over all frequencies:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\star = \int_0^\infty{I(\nu,T) d\nu}\int_{half-sphere}{d\Omega}}

First, the solid angle:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\star = \int_0^\infty{I(\nu,T) d\nu} \int_{half-sphere}{\sin\theta ~ d\theta ~ d\phi} = \int_0^\infty{I(\nu,T) d\nu} \int_0^{2\pi}{d\phi} \int_0^\frac{\pi}{2}{\cos\theta ~ \sin\theta ~ d\theta}}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = 2\pi\int_0^\infty{I(\nu,T) d\nu} \int_0^1{\sin\theta ~ d(\sin\theta)} = \pi\int_0^\infty{I(\nu,T) d\nu}}

Now we substitute Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I(\nu,T)} from Planck's law:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\star = \pi\int_0^\infty{\frac{2 h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{kT}}-1} d\nu} = \frac{2\pi h}{c^2}\int_0^\infty{\frac{\nu^3}{e^{\frac{h\nu}{kT}}-1} d\nu}}

We now define:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \chi \equiv \frac{h\nu}{kT}}

and from this substitute the following:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \nu = \frac{kT}{h}\chi ~, ~~ d\nu = \frac{kT}{h}d\chi}

into our above expression for j^*:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\star = \frac{2\pi h}{c^2}\int_0^\infty{\frac{(\frac{kT}{h}\chi)^3}{e^\chi-1} \frac{kT}{h}d\chi} = \frac{2\pi k^4}{h^3 c^2}T^4\int_0^\infty{\frac{\chi^3}{e^\chi-1} d\chi} }

Using the given integral:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int_0^{\infty}[x^3 /[e^x -1] dx = \pi^4 /15}

we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\star = \frac{2\pi^5 k^4}{15h^3 c^2}T^4}

Identifying the proportionality constant:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma \equiv \frac{2\pi^5 k^4}{15h^3 c^2}}

we now have the Stefan-Boltzmann law:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j^\star = \sigma T^4}

Problem 2

Part a

The de Broglie wavelength of a particle can be written as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda = \frac{h}{p} }

The momentum can be rewritten as:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{p}= m \mathbf{v}\,\! }

with m being the rest mass and v being the velocity.

In order to write the expression for de Broglie wavelength in terms of potential difference V, a relationship between p and v must be established. This can be done by equating two equations for kinetic energy.

The first equation is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \,\! E_k= |qV|}

with q being the charge and V being the potential difference. The next equation is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_k = \begin{matrix} \frac{1}{2} \end{matrix} mv^2 }

Thus, we end up with

$\,\!E_{k}={\begin{matrix}{\frac {1}{2}}\end{matrix}}mv^{2}=|qV|$

Next, we must use a little algebra in order to match our previous equation of p.

$\,\!{\begin{matrix}{\frac {1}{2}}\end{matrix}}mv^{2}=|qV|$

$\,\!(mv)^{2}=2m|qV|$

$\,\!p=mv={\sqrt {2m|qV|}}$

Just sub this into the earlier equation for the de Broglie wavelength and you obtain:

$\,\!\lambda ={\frac {h}{\sqrt {2m|qV|}}}$

To get an expression for the de Broglie wavelength (in nanometers) of an electron in terms of the potential difference V (in volts) through which it has been accelerated, we need to identify some constants.

m_e = 9.10938188 x 10^-31 kg

q = 1.6022 x 10^-19 C

h = 6.626×10³⁴ kg m² s^-1

1 V = 1 kg m² s^-2 C^-1

Therefore,

$\,\!\lambda _{e}={\frac {1.226426nm}{\sqrt {|V/1V|}}}$