Magnetic field in the hydrogen atom with Bohr theory

Question:
Use Bohr theory. Find the magnetic field by electron's exerting at the nucleus of the hydrogen atom. The radius of hydrogen is assumed to be 0.529 $\times$ 10$^{-10}$ m.

Answer:
We can assume that the electron moves around the nucleus as making a circular current. According to the Biot-Savart Law, we have the magnetic field of a circular current as
\[
B_z = \frac{\mu_0 I}{4 \pi}\int^{2\pi r}_{0}\frac{dl}{r^2}=\frac{\mu_0 I}{2r}
\]
Now, we need to obtain the current. The electric current is defined by $I=\frac{Q}{T}$. The charge of an electron or a proton is 1.61 $\times$ 10$^{-19}$ C. In order to get the current, we have to find the time period. That is
\[
T=\frac{2\pi r}{v}
\]
where $v$ is the tangential velocity of the electron. So the current is expressed by
\[
I = \frac{Q}{\frac{2\pi r}{v}}
\]
In terms of the classical theory, the centripetal force is equal to the Coulomb force. Then, solve for the velocity.
\begin{eqnarray}
\frac{mv^2}{r}&=&\frac{ke^2}{r^2} \\
v&=&\sqrt{\frac{ke^2}{mr}}  \\
  &=& \sqrt{\frac{8.99 \times 10^9 (1.61 \times 10^{-19})^2}{9.11 \times 10^{-31}\cdot 0.529 \times 10^{-10}}}  \\
  &=& 2.20 \times 10^6 \ \mathrm{m/s}
\end{eqnarray}
Therefore, we calculate the current.
\[
I=\frac{Q}{\frac{2\pi r}{v}}=\frac{1.61 \times 10^{-19}}{\frac{2\pi \times 0.529\times 10^{-10}}{2.20\times 10^6}}=1.07\times 10^{-3} \ \mathrm{A}
\]
Then, we can find the magnetic field.
\[
B=\frac{\mu_0 I}{2r}=\frac{4\pi \times 10^{-7} \times1.07\times 10^{-3}}{2 \times 0.529\times 10^{-10}}= 12.7 \ \mathrm{T}
\]

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Uncertainty principle: Time and energy with Delta baryon

Question:
The Delta baryon, $\Delta$, is known as the lowest nucleon resonance that has a mass of 1232 MeV/c$^2$ and a width of 120 MeV/c$^2$. Its spin and isospin are equally 3/2. Find the lifetime of this particle.

Answer:
According to the uncertainty principle between time and energy, we have
\[
\Delta E \cdot \Delta t \geqq \frac{\hbar}{2}
\]
The multiplication of uncertainties is greater than or equal to a half of the reduced Planck constant (or Dirac's constant). Solve for the time and plug in numbers. In this unit system, the constant $\hbar$ should be $\hbar \times c$ = 197.33 MeV fm, where fm is 10$^{-15}$ m. Therefore,
\[
\Delta t \sim \frac{\hbar c}{2\Delta E c} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{MeV \ fm}{MeV / c}} = \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{fm}{c}}  \\
= \frac{197.33}{2 \cdot 120} \ \mathrm{\frac{10^{-15}m}{3.00 \times 10^8 m/s}} \\
= 2.74 \times 10^{-24} \ \mathrm{s}
\]

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Hydrogen molecules: Boltzmann factor

Question:
There are hydrogen molecules in a closed system by contacting a heat bath at $T=300$ K. Find the ratio of hydrogen molecules in the first rotational energy level relative to the ground state. (The molecular distance is given as $r=1.06$ angstroms.)

Answer:
This is a canonical ensemble in the sense of statistical mechanics. This gives a probability to each distinct state in terms of the Boltzmann constant.
\[
P=e^{-\frac{E_i}{kT}}
\]
The rotational part of the Schroedinger equation is
\begin{equation}
\frac{-\hbar^2}{2I}\left[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta \frac{\partial}{\partial\theta}\right)+\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2}\right]\psi=E\psi
\end{equation}
where $I$ is the moment of inertia, $\mu r^2$. The rotational energy level is
\begin{equation}
E_j = \frac{\hbar^2}{2\mu r^2}j(j+1)
\end{equation}
The hydrogen atom has a proton and an electron. Thus, we have the mass.
\[
m_H = 1.67262\times 10^{-27} + 9.10938\times 10^{-31} = 1.67353\times 10^{-27} \ \mathrm{kg}
\]
The reduced mass for the molecule becomes
\[
\mu = \frac{m_{H1} m_{H2}}{m_{H1}+m_{H2}}=\frac{m_H}{2}=8.36765\times 10^{-28} \ \mathrm{kg}
\]
because $m_{H1}=m_{H2}$. The relative distance between two molecules is $1.06 \times 10^{-10}$ m. The ground state of energy is obviously $E_0 = 0$ from (1). Then, calculate the first level.
\[
E_1=\frac{\hbar^2}{2\mu r^2}2=\frac{(1.05457\times 10^{-34})^2}{8.36765\times 10^{-28} (1.06 \times 10^{-10})^2}=1.18287\times 10^{-21} \ \mathrm{J}
\]
Now, calculate the following:
\[
kT = 1.38064 \times 10^{-23} \times 300 = 4.14192 \times 10^{-21} \ \mathrm{J}
\]
Therefore, the probability is
\[
P = \exp\left[-\frac{E_1}{kT}\right] = \exp\left[-\frac{1.18287\times 10^{-21}}{4.14192 \times 10^{-21}}\right] = 0.752 = 75.2 \%
\]

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Proton, neutron, and pi mesons

Question:
Consider a proton, a neutron, and pi mesons. These are also called hadrons which are constructed with quarks. Particularly, these particles consist of up and down quarks. Confirm the structure of each particle and its charge.

Answer:
The up quark has $+\frac{2}{3}e$ as the electric charge. The down quark has $-\frac{1}{3}e$. Protons and neutrons are specifically called baryons which consist of three quarks. A proton contains uud quarks. The charge will be
\[
\mathrm{charge \ of \ proton} = \frac{2}{3}e+\frac{2}{3}e-\frac{1}{3}e = +e
\]
which is experimentally proven.
Likewise, we can find the charge of a neutron (udd).
\[
\mathrm{charge \ of \ neutron} = \frac{2}{3}e-\frac{1}{3}e-\frac{1}{3}e = 0
\]
which is consistent with the experimental result.
Now, mesons are made of a quark and an anti-quark, and note that an anti-quark gets the opposite charge to its particle. The pion has three states:
\begin{eqnarray*}
\pi^+ &=& u\overline{d} = \frac{2}{3}e+\frac{1}{3}e = +e  \\
\pi^- &=& \overline{u}d = -\frac{2}{3}e-\frac{1}{3}e = -e  \\
\pi^0 &=& \frac{u\overline{u} - d\overline{d}}{\sqrt{2}} = 0
\end{eqnarray*}
Notice that $\pi^-$ is the anti-particle of $\pi^+$.

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Work function

Question:
450-nm (450$\times$10$^{-9}$ m) wavelength light is incident on sodium surface for which the threshold wavelength of the photoelectrons is 542 nm (542$\times$10$^{-9}$ m). Find the work function of sodium and kinetic energy of the incident light after photoelectrons released.

Answer:
The work function is defined as the minimum energy to obtain electrons from the material surface to infinite distance. The frequency of some light is $\nu$. The partial energy of light may be used for the minimum threshold, which is the work function, $\phi$. The extra energy becomes the kinetic energy, $T$. Namely, we have the following relationship:
\begin{equation}
h\nu = \phi + T
\end{equation}
The Planck's constant, $h$, is 6.63 $\times$ 10$^{-34}$ Js. The frequency, $\nu$, is given by the speed of light, $c$, divided by wavelength, $\lambda$. In this case, the threshold wavelength is given, so we can exclude the term of the kinetic energy.
\begin{eqnarray}
\phi &=& h\nu = \frac{hc}{\lambda}  \\
       &=& \frac{6.63\times 10^{-34}\cdot 3.00\times 10^8}{542\times 10^{-9}}  \\
      &=& 3.67 \times 10^{-19} \ \mathrm{J}
\end{eqnarray}
One joule is 6.24 $\times$ 10$^{18}$ electronvolts. Thus, we can convert it into eV.
\[
\phi = 3.67 \times 10^{-19} \times 6.24 \times 10^{18} = 2.29 \ \mathrm{eV}
\]
If we use 450-nm light, we have the total energy as
\[
h\nu' = \frac{6.63\times 10^{-34}\cdot 3.00\times 10^8}{450\times 10^{-9}}=4.42\times 10^{-19} \ \mathrm{J}
\]
Namely, it is 2.76 eV. Using equation (1), we have the kinetic energy.
\[
T = h\nu' - \phi = 2.76 - 2.29 = 0.47 \ \mathrm{eV}
\]

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Infinite square well potential

Question:
There is an infinite potential well of width $a$. Find the ground state energy.

Answer:
The potential energies for each domain are
\begin{eqnarray}
V(x) &=& \infty \quad & &(x \leqq 0, \ x \geqq a)  \\
V(x) &=& 0 \quad & &(0 < x <  a)
\end{eqnarray}
For $V(x) = \infty$, obviously there is no particle since it gives zero for the wave function. For the other domain, the hamiltonian is
\[
H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}
\]
Therefore the Schroedinger equation becomes
\begin{eqnarray}
-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} &=& E\psi   \\
\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi &=& 0  \\
\frac{d^2\psi}{dx^2} + k^2\psi &=& 0
\end{eqnarray}
where $k^2=\frac{2mE}{\hbar^2}$.
This differential equation has the general solution.
\begin{equation}
\psi(x) = A\sin kx + B\cos kx
\end{equation}
The wave function is a continuous function; and both edges must be zero due to the infinite potential.
\begin{equation}
\psi(0) = \psi(a) = 0
\end{equation}
This indicates that the wave function must be odd because $0<x<a$. Thus, from equation (6), only the sine function survives, so the wave function is $\psi(x)=A\sin kx$. From (7),
\begin{equation}
A\sin ka = 0
\end{equation}
This gives
\begin{equation}
k_n a = n\pi, \quad n=0,1,2,...
\end{equation}
Since $k=\frac{n\pi}{a}$, the wave function becomes
\begin{equation}
\psi(x) = A\sin \left(\frac{n\pi x}{a}\right)
\end{equation}
Let us now normalize the function.
\begin{eqnarray}
A^2\int^a_0 \sin^2\left(\frac{n\pi x}{a}\right) dx = 1
\end{eqnarray}
Put $\theta=\frac{n\pi x}{a}$, so $dx=\frac{a}{n\pi}d\theta$ and $0<\theta<n\pi$.
\begin{eqnarray}
& &\frac{A^2a}{n\pi}\int^{n\pi}_0 \sin^2\theta d\theta = 1  \\
& &\frac{A^2a}{n\pi}\left[\frac{\theta}{2}-\frac{\sin 2\theta}{4}\right]^{n\pi}_0 = 1   \\
& &\frac{A^2a}{n\pi}\frac{n\pi}{2} = 1   \\
& &A = \sqrt{\frac{2}{a}}
\end{eqnarray}
Hence, the wave function is completed as
\begin{equation}
\psi(x) = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)
\end{equation}
Now, we obtain the energy eigen values. From equations $k^2=\frac{2mE}{\hbar^2}$ and $k=\frac{n\pi}{a}$, we have
\begin{equation}
E_n = \frac{\hbar^2 k^2_n}{2m} = \frac{\hbar^2 n^2 \pi^2}{2ma^2}
\end{equation}
$n=0$ corresponds to $E=0$, so the ground state energy is when $n=1$.

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One dimensional quantum system: Step potential

Question:
Find the coefficients of transmission and reflection for a particle incident on the potential shown. The particle has energy, $E$, lager than the potential energy, $V_0$.

Answer:
This is a quantum system, so we can employ the Schroedinger equation.
\[
H\psi = E\psi
\]
where $H$, $\psi$, and $E$ are the hamiltonian, the wave function, and the energy eigen value, respectively. The hamiltonian operator in general consists of the kinetic and potential energies. In this case, the incident beam is a free particle, so there is zero potential energy before incident on the potential. Thus, we need to consider two cases of the equation:
When $x<0$,
\begin{eqnarray}
-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \psi = E\psi   \\
\psi'' + \frac{2mE}{\hbar^2} \psi = 0
\end{eqnarray}
The momentum can be expressed as $p=\hbar k$ where $k$ is the wave number. Thus,
\[
E = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m}
\]
We can derive
\[
k^2 = \frac{2mE}{\hbar^2}
\]
The equation (2) will become
\[
\psi'' + k^2 \psi = 0
\]
The solution is
\begin{equation}
\psi_I = Ae^{ikx} + Be^{-ikx}
\end{equation}
When $x>0$,
\begin{eqnarray}
-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \psi = (E-V_0)\psi   \\
\psi'' + \frac{2m(E-V_0)}{\hbar^2} \psi = 0
\end{eqnarray}
Let us put $k'=\frac{2m(E-V_0)}{\hbar^2}$. The solution is
\begin{equation}
\psi_{II} = Ce^{ik'x} + De^{-ik'x}
\end{equation}
The term $De^{-ik'x}$ represents the wave emanating from the right side, which is none. Therefore, we can let $D=0$. Namely,
\begin{equation}
\psi_{II} = Ce^{ik'x}
\end{equation}
Due to the consistency of quantum theory, the wave function and its first derivative must be continuous at $x=0$ in this problem. This means
\begin{eqnarray}
\psi_I(0) &=& \psi_{II}(0)  \\
\frac{d}{dx}\psi_I(0) &=& \frac{d}{dx}\psi_{II}(0)
\end{eqnarray}
Use equations (3) and (7).
\begin{eqnarray}
A + B &=& C  \\
A - B &=& \frac{k'}{k}C
\end{eqnarray}
Take the ratios $C/A$, which is related to the transmission, and $B/A$, which is related to the reflection.
\begin{equation}
\frac{C}{A}=\frac{2}{1+k'/k}, \quad \frac{B}{A}=\frac{1-k'/k}{1+k'/k}
\end{equation}
Now, the current density $J$ for the Schroedinger equation can be derived from the continuity equation, $\frac{\partial \rho}{\partial t}+\frac{\partial J}{\partial x}=0$:
\begin{equation}
J = \frac{\hbar^2}{2mi}\left(\psi^* \frac{d\psi}{dx} - \psi \frac{d\psi^*}{dx}\right)
\end{equation}
We can obtain the current density for the incident, transmission, and reflection by using equation (13); and $\psi_{\mathrm{inci}}=Ae^{ik}$, $\psi_{\mathrm{trans}}=Ce^{ik'}$, and $\psi_{\mathrm{reflec}}=Be^{ik}$
\begin{eqnarray*}
J_{\mathrm{inci}} &=& \frac{\hbar^2}{2mi}2ik|A|^2  \\
J_{\mathrm{trans}} &=& \frac{\hbar^2}{2mi}2ik'|C|^2  \\
J_{\mathrm{reflec}} &=& \frac{\hbar^2}{2mi}2ik|B|^2
\end{eqnarray*}
Let us define transmission and reflection coefficients, $T$ and $R$.
\begin{eqnarray*}
T &\equiv& \left| \frac{J_{\mathrm{trans}}}{J_{\mathrm{inci}}} \right|  = \left|\frac{C}{A}\right|^2 \frac{k'}{k}\\
R &\equiv& \left| \frac{J_{\mathrm{reflec}}}{J_{\mathrm{inci}}} \right| = \left|\frac{B}{A}\right|^2
\end{eqnarray*}
Use the result in (12); then, we have
\begin{equation*}
T=\frac{4k'/k}{(1+k'/k)^2}, \quad R=\left|\frac{1-k'/k}{1+k'/k}\right|^2
\end{equation*}

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Momentum space to configuration space

Question:
In quantum theory, the wave function can be convertible either in momentum space or configuration space. We have a wave function in momentum space: $\phi(p) = N/(p+\alpha^2)$. Find the equivalent wave function in configuration space.

Answer:
We can use the Fourier transformation to obtain the wave function in configuration space.
\begin{eqnarray*}
\Phi(x) &=& \frac{1}{\sqrt{2\pi\hbar}}\int^{\infty}_{-\infty}dp \frac{N}{p^2+\alpha^2}e^{ipx/\hbar}  \\
&=&  \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{0}_{-\infty} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right)  \\
&=&  \frac{N}{\sqrt{2\pi\hbar}}\left(\int^{\infty}_{0} \frac{e^{-ipx/\hbar}}{p^2+\alpha^2}dp+\int^{\infty}_{0} \frac{e^{ipx/\hbar}}{p^2+\alpha^2}dp\right)  \\
&=&  \frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{e^{ipx/\hbar}+e^{-ipx/\hbar}}{p^2+\alpha^2}dp
\end{eqnarray*}
Use the following formula.
\[ \cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} \]
We can rewrite it as
\[ \Phi(x) = 2\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos(px/\hbar)}{p^2+\alpha^2}dp \]
Replace the momentum with the wavenumber; namely, $p=k\hbar$.
\begin{eqnarray*}
\Phi(x) &=& \frac{2}{\hbar}\frac{N}{\sqrt{2\pi\hbar}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk    \\
   &=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk
\end{eqnarray*}
Let us put $\sqrt{\frac{2N^2}{\pi \hbar^3}}$ as $A$. Now, integrate the above in the complex plane. Consider the path of half circle in the upper half plane. The integral becomes
\[ I = A\oint \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk
     = A\left( \int^r_{-r} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk + \int_{C_r}\frac{e^{izx}}{z^2+\alpha^2/\hbar^2}dz \right) \]
where $z$ is the complex variable of $k$.
The second term will become zero because $1/(z^2+\alpha^2/\hbar^2) = 0$ as $|z|\rightarrow \infty$. Due to Jordan's lemma, when $r \rightarrow \infty$, the integral becomes zero. Namely, only the first term survives.
\[ I = A \int^{\infty}_{-\infty} \frac{e^{ikx}}{k^2+\alpha^2/\hbar^2}dk \]
The pole inside the contour is $i\frac{\alpha}{\hbar}$. Using the residue theorem, we obtain
\[ \mathrm{Res} [f, i\alpha/\hbar] = \lim_{z\rightarrow i\frac{\alpha}{\hbar}}\frac{e^{izx}}{z+i\frac{\alpha}{\hbar}} =  \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\]
Thus
\[ I = A \left(2\pi i \frac{e^{-\frac{\alpha}{\hbar}x}}{2i\frac{\alpha}{\hbar}}\right) = A \frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \]
Take the real part of the integral, $I$.
\begin{eqnarray*}
A \int^{\infty}_{0} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk &=& \frac{A}{2}\int^{\infty}_{-\infty} \frac{\cos kx}{k^2+\alpha^2/\hbar^2}dk     \\
   &=& \frac{A}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar}
\end{eqnarray*}
Therefore, we finally have the wave function in configuration space:
\begin{eqnarray*}
\Phi(x) &=& \sqrt{\frac{2N^2}{\pi \hbar^3}}\frac{1}{2}\frac{\pi e^{-(\alpha/\hbar) x}}{\alpha/\hbar} \\
       &=& \frac{N}{\alpha}\sqrt{\frac{\pi}{2\hbar}}e^{-\frac{\alpha}{\hbar}x}
\end{eqnarray*}

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Mechanical model of the proton (angular momentum etc.)

Question:
The proton's mass and radius are 1.67 $\times$ 10$^{-27}$ kg and 1.00 $\times$ 10$^{-15}$ m. This system is governed by quantum and classical regimes. Namely, the proton is assumed to have a rotational uniform solid spherical body and a half spin of quantum angular momentum. Find the equatorial velocity of the proton.

Answer:
In terms of classical sense, the moment of inertia of proton is
\[ I = \frac{2}{5}mr^2 \]
Therefore, the angular momentum becomes
\[ L = I\omega = \frac{2}{5}mr^2\omega \]
For the quantum perspective, the spin angular momentum is given as
\[ S = \frac{1}{2}\hbar \]
Therefore,
\[ \frac{2}{5}mr^2\omega  = \frac{1}{2}\hbar \]
Solve for the angular velocity.
\[ \omega = \frac{1}{2}\hbar \frac{5}{2mr^2} \]
The equatorial velocity is given by
\[ v = r\omega =  \frac{5\hbar}{4mr} \\
      = \frac{5}{4}\frac{1.055\times 10^{-34}\mathrm{Js}}{1.673\times 10^{-27}\mathrm{kg}\times 1.00\times 10^{-15}\mathrm{m}} \]
Then we have
\[ v = 7.88 \times 10^7 \mathrm{m/s} \]

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Normalization of a wave function

Question:
The wave function for the one dimensional harmonic oscillator with the potential energy, $\frac{1}{2}kx^2$, is given as
\[ \Phi_0 = C\exp(-ax^2) \]
Find the constant $C$ when the wave function is normalized. Note that you can use the normalized
Gaussian distribution: $\frac{1}{\sqrt{2\pi}\sigma}\int^{\infty}_{-\infty}\exp(-\frac{x^2}{2\sigma^2})dx = 1$.

Answer:
For the normalized wave function, it has to satisfy:
\[  \int^{\infty}_{-\infty}\Phi_0\Phi^*_0 dx = C^2\int^{\infty}_{-\infty}\exp(-2ax^2) dx = 1\]
where $\Phi^*_0$ is the complex conjugate of $\Phi_0$. Compare this with the Gaussian normal distribution.
\[ 2a = \frac{1}{2\sigma^2}  \\
   \sigma = \frac{1}{\sqrt{4a}}\]
Plug it in the formula of Gaussian distribution.
\[ \frac{1}{\sqrt{2\pi}(1/\sqrt{4a})}\int^{\infty}_{-\infty}\exp(-\frac{x^2}{2(1/4a)})dx = 1 \]
We can see
\[ C^2 = \frac{1}{\sqrt{2\pi}(1/\sqrt{4a})} = \frac{1}{\sqrt{2\pi}}\sqrt{4a} \\
           = \sqrt{\frac{2a}{\pi}} \]
Therefore,
\[ C = (2a/\pi)^{1/4}\]

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