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 generic 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,

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

Part b

$\,\!\lambda _{e}={\frac {1.226426nm}{\sqrt {|3.0x10^{4}V/1V|}}}$

$\,\!\lambda _{e}=7.1x10^{-3}nm$

Part c

$\lambda ={\frac {h}{mv}}$

For this calculation, average values for walking speed (2 m/s) and mass (75 kg) will be used.

$\lambda ={\frac {6.626x10^{-34}}{(75)(2)}}=4.4x10^{-27}nm$

This wavelength is extremely short, suggesting that there is very little wavelike motion when the average person is walking.

Problem 3

Part a

$\,\!E(R)=-{\frac {A}{R^{2}}}+{\frac {B}{R^{10}}}$

In order to solve for A and B, another equation must be obtained. This can be done by taking dE/dR and setting it equal to zero.

$\,\!0={\frac {dE}{dR}}={\frac {2A}{R^{3}}}-{\frac {10B}{R^{11}}}$

Then, solve for B in terms of A:

$\,\!B={\frac {R^{8}}{5}}A=1.312x10^{-5}nm^{8}A$

By plugging in the value of B in the original equation, as well as substituting in all the values given in the problem we obtain:

$\,\!-4eV=-{\frac {A}{(.3nm)^{2}}}+{\frac {(1.312x10^{-5}nm^{8}A)}{(.3nm)^{10}}}$

$\,\!-4eV=-A(8.88nm^{-2})$

$\,\!A=.45nm^{2}eV$

Then plug the value for A back into the equation for B:

$\,\!B=1.312x10^{-5}nm^{8}(.45nm^{2}eV)$

$\,\!B=5.9049x10^{-6}nm^{10}eV$

Problem 4

Part a

The simplest of all atomic systems, the hydrogen atom consists of one proton and one electron. It is the first of a class of atomic systems that contain only one electron. The energy levels $E_{n}$ of such a system are given by:

$E_{n}={\frac {Z^{2}}{n^{2}}}E_{1}$

where $Z$ is the number of protons in the nucleus (the nuclear charge, or the atom's atomic number) and $E_{1}$ is the ground state energy of the hydrogen atom. The latter may be calculated via the Bohr model of the atom:

$E_{1}=-{\frac {m_{e}q_{e}^{4}}{8h^{2}\epsilon _{0}^{2}}}=-13.6\ \mathrm {eV}$

where $m_{e}$ and $q_{e}$ are, respectively, the mass and charge of the electron, $h$ is Planck's constant, and $\epsilon _{0}$ is the vacuum permittivity.

If we place two hydrogen atoms far enough apart so that their interaction is negligible, the total energy of the system is just that of two hydrogen atoms:

$E_{r\rightarrow \infty }=2E_{1}=2(-13.6\ \mathrm {eV} )=-27.2\ \mathrm {eV}$

According to quantum mechanics, the spatial wavefunction for the system can be either symmetric (denoted $\Psi _{S}$) or antisymmetric ($\Psi _{A}$) about the midpoint between the atoms:

If the atoms are far apart, either of these two possibilities results in exactly the same total energy for the system (-27.2 eV). However, as the two atoms are brought closer together, each atom's contribution to the total wavefunction will either add constructively (in the case of $\Psi _{S}$) or destructively (in the case of $\Psi _{A}$), resulting in two substantially different outcomes:

As the above diagram shows, the constructive interference between the two atom's wavefunctions in the symmetric case of $\Psi _{S}$ result in a non-zero value of the wavefunction at the midpoint between the two atoms, and thus the electron probability density $|\Psi _{S}|^{2}$ between the two protons in the symmetric case is also non-zero. Thus the electrons exist between the two protons, interposing their negative charge between the repulsive positive charges of the protons and so holding the molecule together. This electronic state is stable, and is known as the bonding orbital.

In contrast, the two atom's wavefunctions in the antisymmetric case of $\Psi _{A}$ cancel out entirely at the midpoint between the atoms, resulting in an electron probability density $|\Psi _{A}|^{2}$ of zero between the two protons. The electrons therefore do not exist between the two protons, and so they do little to shield their positive charges from one another. The resulting configuration is not stable, and this electronic state is known as the antibonding orbital.