di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Electromagnetism

Electrostatic lens

Consider a uniformly charged metallic ring of radius $R$ and total charge $q$. The ring is a hollow toroid of thickness $2 a \ll R$. This thickness can be neglected in parts A, B, C, and E. The $x y$ plane coincides with the plane of the ring, while the $z$-axis is perpendicular to it, as shown in Figure 1. In parts A and B you might need to use the formula (Taylor expansion)

$$
(1+x)^{\varepsilon} \approx 1+\varepsilon x+\frac{1}{2} \varepsilon(\varepsilon-1) x^{2}, \text { when }|x| \ll 1 \text {. }
$$

<img_4405>

Figure 1. A charged ring of radius $R$.

Part A. Electrostatic potential on the axis of the ring (1 point)
Context question:
A.1 Calculate the electrostatic potential $\Phi(z)$ along the axis of the ring at a $z$ distance from its center (point A in Figure 1).
Context answer:
\boxed{$\Phi(z)=\frac{q}{4 \pi \varepsilon_{0}} \frac{1}{\sqrt{R^{2}+z^{2}}}$.}


Context question:
A.2 Calculate the electrostatic potential $\Phi(z)$ to the lowest non-zero power of $z$, assuming $z \ll R$.
Context answer:
\boxed{$\Phi(z) \approx \frac{q}{4 \pi \varepsilon_{0} R}(1-\frac{z^{2}}{2 R^{2}})$}


Context question:
A.3 An electron (mass $m$ and charge $-e$ ) is placed at point A (Figure 1, $z \ll R$ ). What is the force acting on the electron? Looking at the expression of the force, determine the sign of $q$ so that the resulting motion would correspond to oscillations. The moving electron does not influence the charge distribution on the ring.
Context answer:
$F(z)=-\frac{q e}{4 \pi \varepsilon_{0} R^{3}} z$, q>0


Context question:
A.4 What is the angular frequency $\omega$ of such harmonic oscillations?
Context answer:
\boxed{$\omega=\sqrt{\frac{q e}{4 \pi m \varepsilon_{0} R^{3}}}$}


Extra Supplementary Reading Materials:

Part B. Electrostatic potential in the plane of the ring

In this part of the problem you will have to analyze the potential $\Phi(r)$ in the plane of the ring $(z=0)$ for $r \ll R$ (point $\mathrm{B}$ in Figure 1). To the lowest non-zero power of $r$ the electrostatic potential is given by $\Phi(r) \approx q\left(\alpha+\beta r^{2}\right)$.
Context question:
B.1 Find the expression for $\beta$. You might need to use the Taylor expansion formula given above.
Context answer:
\boxed{$\beta=\frac{1}{16 \pi \varepsilon_{0} R^{3}}$}

Question: B.2 An electron is placed at point $\mathrm{B}$ (Figure $1, r \ll R$ ). What is the force acting on the electron?

Solution:
The potential of the electron is $V(r)=-e \Phi(r)$. Force acting on the electron in the $x y$ plane is

$$
F(r)=-\frac{\mathrm{d} V(r)}{\mathrm{d} r}=+e \frac{\mathrm{d} \Phi(r)}{\mathrm{d} r}=\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r
$$

To have oscilations we need the force to be negative for $r>0$. Thus, $q<0$.



Final answer: $F(r)=+\frac{q e}{8 \pi \varepsilon_{0} R^{3}} r $
Subfield: Electromagnetism


Question: D.2 Determine the dependence of the charge on the ring as a function of time, $q(t)$. $t=0$ corresponds to a time moment when electrons are in the plane of the ring. What is the charge on the ring $q_{0}$ when the absolute value $q(t)$ is maximum? The capacitance of the ring is $C$ (i.e., you do not have to use the actual expression found in D.1).

Remark: the drawn polarity in Figure 3 is for indicative purposes only. The sign should be chosen so that the lens is focusing.

Solution:
Let $q(t)$ be the charge on the ring at a time $t$. Potential of the disk is thus $q(t) / C$. Voltage drop of the resistor is $R_{0} I(t)=R_{0} \mathrm{~d} q / \mathrm{d} t$. Therefore for time $-\frac{d}{2 v}<t<\frac{d}{2 v}$ :

$$
\frac{q(t)}{C}+R_{0} \frac{\mathrm{d} q}{\mathrm{~d} t}=V_{0}
$$

Integrating this equation and keeping in mind that $q(t)=0$ at $t=-d /(2 v)$, we get:

$$
q(t)=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{2 v R_{0} C}} \mathrm{e}^{-\frac{t}{R_{0} C}}\right) .
$$

The charge attains the largest absolute value at $t=d /(2 v)$. The value of the charge at this time is:

$$
q_{0}=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{v R_{0} C}}\right)
$$

When $t>\frac{d}{2 v}$, we get:

$$
\frac{q(t)}{C}+R_{0} \frac{\mathrm{d} q}{\mathrm{~d} t}=0
$$

From here:

$$
q(t)=q_{0} \mathrm{e}^{-\frac{t}{R_{0} C}+\frac{d}{2 v R_{0 C}}}=C V_{0}\left(\mathrm{e}^{\frac{d}{2 v R_{0} C}}-\mathrm{e}^{-\frac{d}{2 v R_{0} C}}\right) \mathrm{e}^{-\frac{t}{R C}} .
$$

Therefore, we obtain:

$$
q(t)= \begin{cases}0 & \text { for } t<-\frac{d}{2 v} ; \\ C V_{0}\left(1-\mathrm{e}^{-\frac{d}{2 v R_{0} C}} \mathrm{e}^{-\frac{t}{R_{0} C}}\right) & \text { for }-\frac{d}{2 v}<t<\frac{d}{2 v} ; \\ C V_{0}\left(\mathrm{e}^{\frac{d}{2 v R_{0} C}}-\mathrm{e}^{-\frac{d}{2 v R_{0} C}}\right) \mathrm{e}^{-\frac{t}{R_{0} C}} & \text { for } t>\frac{d}{2 v} .\end{cases}
$$

For a lens to be focusing we require that charge is negative, therefore $V_{0}<0$. The dependence of charge on time is shown in Figure 6.



![](https://cdn.mathpix.com/cropped/2023_12_21_d0d67f5b5936c04c5e29g-1.jpg?height=582&width=1006&top_left_y=477&top_left_x=525)

Figure 6: Charge on the ring as a function of time.

Final answer: For $-\frac{d}{2 v}<t<\frac{d}{2 v}, \quad q(t)=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{2 v R_{0} C}} \mathrm{e}^{-\frac{t}{R_{0} C}}\right)$.

For $t>\frac{d}{2 v}, \quad q(t)=C V_{0}\left(\mathrm{e}^{\frac{d}{2 v R_{0} C}}-\mathrm{e}^{-\frac{d}{2 v R_{0} C}}\right) \mathrm{e}^{-\frac{t}{R_{0} C}}$.

$q_{0}=C V_{0}\left(1-\mathrm{e}^{-\frac{d}{v R_{0} C}}\right) . \quad$
Subfield: Modern Physics

Particles and Waves

Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.

 Part A. Quantum particle in a box

Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by

$$
V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
\tag{1}
$$

While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.

Question: A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.

Solution:
The width of the potential well $(L)$ should be equal to the half of the wavelength of the de Broglie standing wave $\lambda_{\mathrm{dB}}=h / p$, here $h$ is the Planck's constant and $p$ is the momentum of the particle. Thus $p=h / \lambda_{\mathrm{dB}}=h /(2 L)$, and the minimal possible energy of the particle is

$$
E_{\min }=\frac{p^{2}}{2 m}=\frac{h^{2}}{8 m L^{2}}
$$

Final answer: $E_{\min }=\frac{h^{2}}{8 m L^{2}}$
Subfield: Modern Physics

Particles and Waves

Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.

 Part A. Quantum particle in a box

Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by

$$
V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
\tag{1}
$$

While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.
Context question:
A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.
Context answer:
\boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$}


Extra Supplementary Reading Materials:

The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state.

Question: A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).

Solution:
The potential well should fit an integer number of the de Broglie half-wavelengths: $L=\frac{1}{2} \lambda_{\mathrm{dB}}^{(n)} \cdot n$, $n=1,2, \ldots$ Therefore, particle's momentum, corresponding to the de Broglie wavelength $\lambda_{\mathrm{dB}}^{(n)}$ is

$$
p_{n}=\frac{h}{\lambda_{\mathrm{dB}}^{(n)}}=\frac{h n}{2 L}
$$

and the corresponding energy is

$$
E_{n}=\frac{p_{n}^{2}}{2 m}=\frac{h^{2} n^{2}}{8 m L^{2}}, \quad n=1,2,3, \ldots
\tag{1}
$$

Final answer: $E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$
Subfield: Modern Physics

Particles and Waves

Wave-particle duality, which states that each particle can be described as a wave and vice versa, is one of the central concepts of quantum mechanics. In this problem, we will rely on this notion and just a few other basic assumptions to explore a selection of quantum phenomena covering the two distinct types of particles of the microworld-fermions and bosons.

 Part A. Quantum particle in a box

Consider a particle of mass $m$ moving in a one-dimensional potential well, where its potential energy $V(x)$ is given by

$$
V(x)= \begin{cases}0, & 0 \leq x \leq L \\ \infty, & x<0 \text { or } x>L\end{cases}
\tag{1}
$$

While classical particle can move in such a potential having any kinetic energy, for quantum particle only some specific positive discrete energy levels are allowed. In any such allowed state, the particle can be described as a standing de Broglie wave with nodes at the walls.
Context question:
A.1 Determine the minimal possible energy $E_{\min }$ of the quantum particle in the well. Express your answer in terms of $m, L$, and the Planck's constant $h$.
Context answer:
\boxed{$E_{\min }=\frac{h^{2}}{8 m L^{2}}$}


Extra Supplementary Reading Materials:

The particle's state with minimal possible energy is called the ground state, and all the rest allowed states are called excited states. Let us sort all the possible energy values in the increasing order and denote them as $E_{n}$, starting from $E_{1}$ for the ground state.
Context question:
A.2 Find the general expression for the energy $E_{n}$ (here $n=1,2,3, \ldots$ ).
Context answer:
\boxed{$E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$}


Question: A.3 Particle can undergo instantaneous transition from one state to another only by emitting or absorbing a photon of the corresponding energy difference. Find the wavelength $\lambda_{21}$ of the photon emitted during the transition of the particle from the first excited state $\left(E_{2}\right)$ to the ground state $\left(E_{1}\right)$.

Solution:
The energy of the emitted photon, $E=h c / \lambda$ (here $c$ is the speed of light and $\lambda$ is the photon's wavelength) should be equal to the energy difference $\Delta E=E_{2}-E_{1}$, therefore

$$
\lambda_{21}=\frac{h c}{E_{2}-E_{1}}=\frac{8 m c L^{2}}{3 h}
$$

Final answer: $\lambda_{21}=\frac{8 m c L^{2}}{3 h}$
Subfield: Modern Physics


Question: C.3 What is the particle density of the Rb gas $n_{c}$ if the transition takes place at such= a temperature? For the sake of comparison, calculate also the 'ordinary' particle density $n_{0}$ of an ideal gas at the standard temperature and pressure (STP), i.e. $T_{0}=300 \mathrm{~K}$ and $p_{0}=10^{5} \mathrm{~Pa}$. How many times is the 'ordinary' gas denser? You may assume that the mass of the atoms is equal to 87 atomic mass units $\left(m_{\mathrm{amu}}\right)$.

Solution:
Using the answer to the previous question, we express $n_{c}=\left(3 m k_{\mathrm{B}} T_{c}\right)^{3 / 2} / h^{3}$. Equation of state for the ideal gas gives $n_{0}=p /\left(k_{\mathrm{B}} T\right)$. Numerical estimations yield $n_{c} \approx 1.59 \cdot 10^{18} \mathrm{~m}^{-3}$ and $n_{0} / n_{c} \approx 1.5 \cdot 10^{7}$.

Final answer: Expression: $n_{c}=\frac{\left(3 \cdot 87 m_{\mathrm{amu}} k_{\mathrm{B}} T_{c}\right)^{3 / 2}}{h^{3}}$.

Numerical value: $n_{c} \approx 1.59 \cdot 10^{18} \mathrm{~m}^{-3}$.

Expression: $n_{0}=p /\left(k_{\mathrm{B}} T\right)$.

Numerical value: $n_{0} / n_{c} \approx 1.5 \cdot 10^{7}$.
Subfield: Modern Physics


Question: D.3 Derive the behavior of the potential $V(\vec{r})$ along the coordinate axes, i.e., determine the functions $V_{X}(x) \equiv V(x, 0)$ and $V_{Y}(y) \equiv V(0, y)$. Identify the locations of the extrema of $V_{X}(x)$ and $V_{Y}(y)$ as functions of a single argument. As these functions are periodic, please include in your lists only one representative from each family of periodically repeated minima and maxima.

Solution:
We find

$$
V(x, y)=-\alpha E_{0}^{2}\left\{\frac{3}{2}+\cos (k y \sqrt{3})+\cos \left(\frac{3 k x}{2}+\frac{k y \sqrt{3}}{2}\right)+\cos \left(\frac{3 k x}{2}-\frac{k y \sqrt{3}}{2}\right)\right\}
\tag{8}
$$

and deduce

$$
V_{X}(x)=-\alpha E_{0}^{2}\left\{\frac{5}{2}+2 \cos \frac{3 k x}{2}\right\}
\tag{9}
$$

The potential has a simple cosine form, and the origin in an obvious minimum. Its replica appear at multiples of $\Delta x=4 \pi /(3 k)$. In the midpoint between any two minima, e.g. at $x=\Delta x / 2=2 \pi /(3 k)$, the function $V_{X}(x)$ has its maxima.

Concerning the behaviour along the $y$ axis, we have

$$
V_{Y}(y)=-\alpha E_{0}^{2}\left\{\frac{3}{2}+\cos 2 \varphi+2 \cos \varphi\right\}, \quad \varphi=\sqrt{3} k y / 2
\tag{10}
$$

Looking for the extrema, we find the equation

$$
\sin 2 \varphi+\sin \varphi=0
\tag{11}
$$

$\circ \varphi=0$ is the 'deep' minimum - the lattice site;

- $\varphi=\pi$ is the 'shallow' minimum (later shown to be a saddle point of $V(x, y)$ );
- $\varphi=2 \pi / 3$ and $\varphi=4 \pi / 3$ are maxima.

Final answer: $V_{X}(x)=-\alpha E_{0}^{2}\left\{\frac{3}{2}+2 \cos \frac{3 k x}{2}\right\}$.

$V_{Y}(y)=-\alpha E_{0}^{2}\left\{\frac{3}{2}+\cos 2 \varphi+2 \cos \varphi\right\}, \quad$ here $\varphi=\sqrt{3} k y / 2$.

Minimum $(-\mathrm{a})$ of $V_{X}(x): x=0$.

Maximum (-a) of $V_{X}(x): x=\frac{2 \pi}{3 k}$.

Minimum (-a) of $V_{Y}(y): y=0$ ('deep') and $y=\frac{2 \pi}{\sqrt{3} k}$ ('shallow').

Maximum (-a) of $V_{Y}(y): y=\frac{4 \pi}{3 \sqrt{3} k}$ and $y=\frac{8 \pi}{3 \sqrt{3} k}$.
Subfield: Modern Physics

Particles from the Sun ${ }^{1}$ 

Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.

Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.

Note:

(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant

(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant

(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant

A Radiation from the sun :

Question: A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.

Solution:
Stefan's law: $L_{\odot}=\left(4 \pi R_{\odot}^{2}\right)\left(\sigma T_{\mathrm{s}}^{4}\right)$

$$
T_{\mathrm{s}}=\left(\frac{L_{\odot}}{4 \pi R_{\odot}^{2} \sigma}\right)^{1 / 4}=5.76 \times 10^{3} \mathrm{~K}
$$

Final answer: $5.76 \times 10^{3} $ K
Subfield: Modern Physics

Particles from the Sun ${ }^{1}$ 

Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.

Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.

Note:

(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant

(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant

(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant

A Radiation from the sun :
Context question:
A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
Context answer:
\boxed{$5.76 \times 10^{3} $}


Extra Supplementary Reading Materials:

The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by

$$
u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
$$

where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.

Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.

Question: A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$

Solution:
$$
P_{\text {in }}=\int_{0}^{\infty} u(\nu) d \nu=\int_{0}^{\infty} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} \nu^{3} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right) d \nu
$$

Let $x=\frac{h \nu}{k_{\mathrm{B}} T_{\mathrm{s}}}$. Then, $\nu=\frac{k_{\mathrm{B}} T_{\mathrm{s}}}{h} x \quad d \nu=\frac{k_{\mathrm{B}} T_{\mathrm{s}}}{h} d x$.

$$
P_{\mathrm{in}}=\frac{2 \pi h A R_{\odot}^{2}}{c^{2} d_{\odot}^{2}} \frac{\left(k_{\mathrm{B}} T_{\mathrm{s}}\right)^{4}}{h^{4}} \int_{0}^{\infty} x^{3} e^{-x} d x=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \cdot 6=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}
$$

Final answer: $P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$
Subfield: Modern Physics

Particles from the Sun ${ }^{1}$ 

Photons from the surface of the Sun and neutrinos from its core can tell us about solar temperatures and also confirm that the Sun shines because of nuclear reactions.

Throughout this problem, take the mass of the Sun to be $M_{\odot}=2.00 \times 10^{30} \mathrm{~kg}$, its radius, $R_{\odot}=7.00 \times$ $10^{8} \mathrm{~m}$, its luminosity (radiation energy emitted per unit time), $L_{\odot}=3.85 \times 10^{26} \mathrm{~W}$, and the Earth-Sun distance, $d_{\odot}=1.50 \times 10^{11} \mathrm{~m}$.

Note:

(i) $\int x e^{a x} d x=\left(\frac{x}{a}-\frac{1}{a^{2}}\right) e^{a x}+$ constant

(ii) $\int x^{2} e^{a x} d x=\left(\frac{x^{2}}{a}-\frac{2 x}{a^{2}}+\frac{2}{a^{3}}\right) e^{a x}+$ constant

(iii) $\int x^{3} e^{a x} d x=\left(\frac{x^{3}}{a}-\frac{3 x^{2}}{a^{2}}+\frac{6 x}{a^{3}}-\frac{6}{a^{4}}\right) e^{a x}+$ constant

A Radiation from the sun :
Context question:
A.1 Assume that the Sun radiates like a perfect blackbody. Use this fact to calculate the temperature, $T_{\mathrm{s}}$, of the solar surface.
Context answer:
\boxed{$5.76 \times 10^{3} $}


Extra Supplementary Reading Materials:

The spectrum of solar radiation can be approximated well by the Wien distribution law. Accordingly, the solar energy incident on any surface on the Earth per unit time per unit frequency interval, $u(v)$, is given by

$$
u(v)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi h}{c^{2}} v^{3} \exp \left(-h v / k_{\mathrm{B}} T_{\mathrm{s}}\right)
$$

where $v$ is the frequency and $A$ is the area of the surface normal to the direction of the incident radiation.

Now, consider a solar cell which consists of a thin disc of semiconducting material of area, $A$, placed perpendicular to the direction of the Sun's rays.
Context question:
A2 Using the Wien approximation, express the total radiated solar power, $P_{\mathrm{in}}$, incident on the surface of the <br> solar cell, in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{S}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$
Context answer:
\boxed{$P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$}


Question: A3 Express the number of photons, $n_{\gamma}(v)$, per unit time per unit frequency interval incident on the surface of <br> the solar cell in terms of $A, R_{\odot}, d_{\odot}, T_{\mathrm{s}}, v$ and the fundamental constants $c, h, k_{\mathrm{B}}$.

Solution:
$$
\begin{aligned}
n_{\gamma}(\nu) & =\frac{u(\nu)}{h \nu} \\
& =A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right)
\end{aligned}
$$

Final answer: $n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$
Subfield: Mechanics

A The Extremum Principle in Mechanics

Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts).

<img_4336>

Figure 1

Question: A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$.

Solution:
From the principle of Conservation of Mechanical Energy

$$
\begin{gathered}
\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m v_{2}^{2}+V_{0} \\
v_{2}=\left(v_{1}^{2}-\frac{2 V_{0}}{m}\right)^{1 / 2}
\end{gathered}
$$

Final answer: $v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$
Subfield: Mechanics

A The Extremum Principle in Mechanics

Consider a horizontal frictionless $x-y$ plane shown in Fig. 1. It is divided into two regions, I and II, by a line $\mathrm{AB}$ satisfying the equation $x=x_{1}$. The potential energy of a point particle of mass $m$ in region I is $V=0$ while it is $V=V_{0}$ in region II. The particle is sent from the origin $\mathrm{O}$ with speed $v_{1}$ along a line making an angle $\theta_{1}$ with the $x$-axis. It reaches point $\mathrm{P}$ in region II traveling with speed $v_{2}$ along a line that makes an angle $\theta_{2}$ with the $x$-axis. Ignore gravity and relativistic effects in this entire task T-2 (all parts).

<img_4336>

Figure 1
Context question:
A.1 Obtain an expression for $v_{2}$ in terms of $m, v_{1}$ and $V_{0}$.
Context answer:
\boxed{$v_{2}=(v_{1}^{2}-\frac{2 V_{0}}{m})^{1 / 2}$}


Question: A.2 Express $v_{2}$ in terms of $v_{1}, \theta_{1}$ and $\theta_{2}$.

Solution:
At the boundary there is an impulsive force $(\propto d V / d x)$ in the $-x$ direction. Hence only the velocity component in the $x$-direction $v_{1 x}$ suffers change. The component in the $y$-direction remains unchanged. Therefore

$v_{1 y}=v_{2 y}$

$v_{1} \sin \theta_{1}=v_{2} \sin \theta_{2}$

Final answer: $v_2=\frac{v_1 \sin\theta_1}{\sin \theta_2}$