weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: (e) Hence, write down an expression of the motion of $m$ which is a constant.

Solution: From d),

$$
\begin{gathered}
\ddot{x} \dot{x}=2 \omega \dot{x} \dot{y}-\dot{x} \frac{\partial U}{\partial x} \\
\ddot{y} \dot{y}=-2 \omega \dot{x} \dot{y}-\dot{y} \frac{\partial U}{\partial y} \\
\ddot{x} \dot{x}+\ddot{y} \dot{y}=-\dot{x} \frac{\partial U}{\partial x}-\dot{y} \frac{\partial U}{\partial y} \\
\frac{d}{d t}\left[\frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)\right]=-\left(\dot{x} \frac{\partial U}{\partial x}+\dot{y} \frac{\partial U}{\partial y}\right)=-\frac{d U}{d t} \\
\frac{d}{d t}\left[\frac{1}{2}\left(\dot{x}^{2}+\dot{y}^{2}\right)+U\right]=0 \\
\text { Constant }=-2 U-v^{2}
\end{gathered}
$$

Answer: $-2 U-v^{2}$
Problem: a - Calculate the flattening factor, given that the gravitational constant is $6.67 \times 10^{-11}$ N.m $\cdot \mathrm{kg}^{-2}$.

Solution: a - 1st method

For equilibrium we have $\mathrm{F}_{\mathrm{c}}=\mathrm{F}_{\mathrm{g}}+\mathrm{N}$ where $\mathrm{N}$ is normal to the surface.

Resolving into horizontal and vertical components, we find:

<img_4429>

$$
\begin{aligned}
& F_{g} \cdot \cos (\phi)=F_{c}+N \cdot \sin (\alpha) \\
& \quad F_{g} \cdot \sin (\phi)=N \cdot \cos (\alpha)
\end{aligned} \rightarrow F_{g} \cdot \cos (\phi)=F_{c}+F_{g} \cdot \sin (\phi) \cdot \operatorname{tg}(\alpha)
$$

From:

$$
F_{g}=\frac{G \cdot M}{r^{2}}, \quad F_{c}=\omega^{2} \cdot r, x=r \cdot \cos (\phi), y=r \cdot \sin (\phi) \text { en } \operatorname{tg}(\alpha)=\frac{d y}{d x}
$$

we find:

$$
y \cdot d y+\left(1-\frac{\omega^{2} \cdot r^{3}}{G \cdot M}\right) \cdot x \cdot d x=0
$$

where:

$$
\frac{\omega^{2} \cdot r^{3}}{G \cdot M} \approx 7 \cdot 10^{-4}
$$

This means that, although $r$ depends on $x$ and $y$, the change in the factor in front of $x d x$ is so slight that we can take it to be constant. The solution of Eq. (1) is then an ellipse:



$$
\frac{x^{2}}{r_{e}^{2}}+\frac{y^{2}}{r_{p}^{2}}=1 \rightarrow \frac{r_{p}}{r_{e}}=\sqrt{1-\frac{\omega^{2} \cdot r^{3}}{G \cdot M}} \approx 1-\frac{\omega^{2} \cdot r^{3}}{2 \cdot G \cdot M}
$$

and from this it follows that:

$$
\epsilon=\frac{r_{e}-r_{p}}{r_{e}}=\frac{\omega^{2} \cdot r^{3}}{2 \cdot G \cdot M} \approx 3,7 \cdot 10^{-4}
$$

Answer: $3.7 \times 10^{-4}$
Problem: 2.1 Find the ratio $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}$ in terms of $\gamma, P_{a}$ and $R_{0}$.

Solution: The bubble is surrounded by air.

<img_4487>

$P_{a}, T_{a}, \rho_{a}$

Cutting the sphere in half and using the projected area to balance the forces give

$$
\begin{aligned}
P_{i} \pi R_{0}^{2} & =P_{a} \pi R_{0}^{2}+2\left(2 \pi R_{0} \gamma\right) \\
P_{i} & =P_{a}+\frac{4 \gamma}{R_{0}}
\end{aligned}
$$

The pressure and density are related by the ideal gas law:

$P V=n R T$ or $P=\frac{\rho R T}{M}$, where $M=$ the molar mass of air.

Apply the ideal gas law to the air inside and outside the bubble, we get

$$
\begin{aligned}
& \rho_{i} T_{i}=P_{i} \frac{M}{R} \\
& \rho_{a} T_{a}=P_{a} \frac{M}{R}, \\
& \frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}=\frac{P_{i}}{P_{a}}=\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right]
\end{aligned}
$$

Answer: $1+\frac{4 \gamma}{R_{0} P_{a}}$
Problem: 2.2 Find the numerical value of $\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}-1$ using $\gamma=0.0250 \mathrm{Nm}^{-1}, R_{0}=1.00 \mathrm{~cm}$, and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$.

Solution: Using $\gamma=0.025 \mathrm{Nm}^{-1}, R_{0}=1.0 \mathrm{~cm}$ and $P_{a}=1.013 \times 10^{5} \mathrm{Nm}^{-2}$, the numerical value of the ratio is

$$
\frac{\rho_{i} T_{i}}{\rho_{a} T_{a}}=1+\frac{4 \gamma}{R_{0} P_{a}}=1+0.0001
$$

(The effect of the surface tension is very small.)

Answer: 1.0001
Problem: 2.3 The bubble is initially formed with warmer air inside. Find the minimum numerical value of $T_{i}$ such that the bubble can float in still air. Use $T_{a}=300 \mathrm{~K}, \rho_{s}=1000 \mathrm{kgm}^{-3}$, $\rho_{a}=1.30 \mathrm{kgm}^{-3}, t=100 \mathrm{~nm}$ and $g=9.80 \mathrm{~ms}^{-2}$.

Solution: Let $W=$ total weight of the bubble, $F=$ buoyant force due to air around the bubble

$$
\begin{aligned}
W & =(\text { mass of film }+ \text { mass of air }) g \\
& =\left(4 \pi R_{0}^{2} \rho_{s} t+\frac{4}{3} \pi R_{0}^{3} \rho_{i}\right) g \\
& =4 \pi R_{0}^{2} \rho_{s} t g+\frac{4}{3} \pi R_{0}^{3} \frac{\rho_{a} T_{a}}{T_{i}}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right] g
\end{aligned}
$$

The buoyant force due to air around the bubble is

$$
B=\frac{4}{3} \pi R_{0}^{3} \rho_{a} g
$$

If the bubble floats in still air,

$$
\begin{aligned}
B & \geq W \\
\frac{4}{3} \pi R_{0}^{3} \rho_{a} g & \geq 4 \pi R_{0}^{2} \rho_{s} \operatorname{tg}+\frac{4}{3} \pi R_{0}^{3} \frac{\rho_{a} T_{a}}{T_{i}}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right] g
\end{aligned}
$$

Rearranging to give

$$
\begin{aligned}
T_{i} & \geq \frac{R_{0} \rho_{a} T_{a}}{R_{0} \rho_{a}-3 \rho_{s} t}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right] \\
& \geq 307.1 \mathrm{~K}
\end{aligned}
$$

The air inside must be about $7.1^{\circ} \mathrm{C}$ warmer.

Answer: 307.1
Problem: 2.4 Find the minimum velocity $u$ of an updraught (air flowing upwards) that will keep the bubble from falling at thermal equilibrium. Give your answer in terms of $\rho_{s}, R_{0}, g, t$ and the air's coefficient of viscosity $\eta$. You may assume that the velocity is small such that Stokes's law applies, and ignore the change in the radius when the temperature lowers to the equilibrium. The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$.

Solution: Ignore the radius change $\rightarrow$ Radius remains $R_{0}=1.0 \mathrm{~cm}$

(The radius actually decreases by $0.8 \%$ when the temperature decreases from $307.1 \mathrm{~K}$ to $300 \mathrm{~K}$. The film itself also becomes slightly thicker.)

The drag force from Stokes' Law is $F=6 \pi \eta R_{0} u$

If the bubble floats in the updraught,

$F \geq W-B$

$6 \pi \eta R_{0} u \geq\left(4 \pi R_{0}^{2} \rho_{s} t+\frac{4}{3} \pi R_{0}^{3} \rho_{i}\right) g-\frac{4}{3} \pi R_{0}^{3} \rho_{a} g$

When the bubble is in thermal equilibrium $T_{i}=T_{a}$.

$6 \pi \eta R_{0} u \geq\left(4 \pi R_{0}^{2} \rho_{s} t+\frac{4}{3} \pi R_{0}^{3} \rho_{a}\left[1+\frac{4 \gamma}{R_{0} P_{a}}\right]\right) g-\frac{4}{3} \pi R_{0}^{3} \rho_{a} g$

Rearranging to give

$u \geq \frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$

Answer: $\frac{4 R_{0} \rho_{s} t g}{6 \eta}+\frac{\frac{4}{3} R_{0}^{2} \rho_{a} g\left(\frac{4 \gamma}{R_{0} P_{a}}\right)}{6 \eta}$
Problem: 2.5 Calculate the numerical value for $u$ using $\eta=1.8 \times 10^{-5} \mathrm{kgm}^{-1} \mathrm{~s}^{-1}$.

Solution: The numerical value is $u \geq 0.36 \mathrm{~m} / \mathrm{s}$.

Answer: $0.36$
Problem: 2.6 If this spherical bubble is now electrified uniformly with a total charge $q$, find an equation describing the new radius $R_{1}$ in terms of $R_{0}, P_{a}, q$ and the permittivity of free space $\varepsilon_{0}$.

Solution: When the bubble is electrified, the electrical repulsion will cause the bubble to expand in size and thereby raise the buoyant force.

The force/area is (e-field on the surface $\times$ charge/area)


Consider a very thin pill box on the soap surface.

<img_4490>

$E=$ electric field on the film surface that results from all other parts of the soap film, excluding the surface inside the pill box itself.

$E_{q}=$ total field just outside the pill box $=\frac{q}{4 \pi \varepsilon_{0} R_{1}^{2}}=\frac{\sigma}{\varepsilon_{0}}$

$=E+$ electric field from surface charge $\sigma$

$=E+E_{\sigma}$

Using Gauss's Law on the pill box, we have $E_{\sigma}=\frac{\sigma}{2 \varepsilon_{0}}$ perpendicular to the film as a result of symmetry.

Therefore, $E=E_{q}-E_{\sigma}=\frac{\sigma}{\varepsilon_{0}}-\frac{\sigma}{2 \varepsilon_{0}}=\frac{\sigma}{2 \varepsilon_{0}}=\frac{1}{2 \varepsilon_{0}} \frac{q}{4 \pi R_{1}^{2}}$


The repulsive force per unit area of the surface of bubble is

$$
\left(\frac{q}{4 \pi R_{1}^{2}}\right) E=\frac{\left(q / 4 \pi R_{1}^{2}\right)^{2}}{2 \varepsilon_{0}}
$$

Let $P_{i}^{\prime}$ and $\rho_{i}^{\prime}$ be the new pressure and density when the bubble is electrified.

This electric repulsive force will augment the gaseous pressure $P_{i}^{\prime}$.

$P_{i}^{\prime}$ is related to the original $P_{i}$ through the gas law.

$P_{i}^{\prime} \frac{4}{3} \pi R_{1}^{3}=P_{i} \frac{4}{3} \pi R_{0}^{3}$

$P_{i}^{\prime}=\left(\frac{R_{0}}{R_{1}}\right)^{3} P_{i}=\left(\frac{R_{0}}{R_{1}}\right)^{3} P_{a}$

In the last equation, the surface tension term has been ignored.

From balancing the forces on the half-sphere projected area, we have (again ignoring the surface tension term)

$$
\begin{aligned}
& P_{i}^{\prime}+\frac{\left(q / 4 \pi R_{1}^{2}\right)^{2}}{2 \varepsilon_{0}}=P_{a} \\
& P_{a}\left(\frac{R_{0}}{R_{1}}\right)^{3}+\frac{\left(q / 4 \pi R_{1}^{2}\right)^{2}}{2 \varepsilon_{0}}=P_{a}
\end{aligned}
$$



Rearranging to get

$$
\left(\frac{R_{1}}{R_{0}}\right)^{4}-\left(\frac{R_{1}}{R_{0}}\right)-\frac{q^{2}}{32 \pi^{2} \varepsilon_{0} R_{0}^{4} P_{a}}=0
\tag{17}
$$

Note that (17) yields $\frac{R_{1}}{R_{0}}=1$ when $q=0$, as expected.

Answer: $\left(\frac{R_{1}}{R_{0}}\right)^{4}-\left(\frac{R_{1}}{R_{0}}\right)-\frac{q^{2}}{32 \pi^{2} \varepsilon_{0} R_{0}^{4} P_{a}}=0$
Problem: 2.7 Assume that the total charge is not too large (i.e. $\frac{q^{2}}{\varepsilon_{0} R_{0}^{4}}<<P_{a}$ ) and the bubble only experiences a small increase in its radius, find $\Delta R$ where $R_{1}=R_{0}+\Delta R$.

Given that $(1+x)^{n} \approx 1+n x$ where $x \ll 1$.

Solution: Approximate solution for $R_{1}$ when $\frac{q^{2}}{32 \pi^{2} \varepsilon_{0} R_{0}^{4} P_{a}}<<1$

Write $R_{1}=R_{0}+\Delta R, \Delta R<<R_{0}$

Therefore, $\frac{R_{1}}{R_{0}}=1+\frac{\Delta R}{R_{0}},\left(\frac{R_{1}}{R_{0}}\right)^{4} \approx 1+4 \frac{\Delta R}{R_{0}}$

Eq. (17) gives:

$$
\begin{aligned}
& \Delta R \approx \frac{q^{2}}{96 \pi^{2} \varepsilon_{0} R_{0}^{3} P_{a}} \\
& R_{1} \approx R_{0}+\frac{q^{2}}{96 \pi^{2} \varepsilon_{0} R_{0}^{3} P_{a}} \approx R_{0}\left(1+\frac{q^{2}}{96 \pi^{2} \varepsilon_{0} R_{0}^{4} P_{a}}\right)
\end{aligned}
$$

Answer: $\frac{q^{2}}{96 \pi^{2} \varepsilon_{0} R_{0}^{3} P_{a}}$
Problem: 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

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

Answer: $E_{\min }=\frac{h^{2}}{8 m L^{2}}$
Problem: 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

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

Answer: $E_{n}=\frac{h^{2} n^{2}}{8 m L^{2}}$
Problem: 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

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

Answer: $\lambda_{21}=\frac{8 m c L^{2}}{3 h}$
Problem: C.1 Given a non-interacting gas of ${ }^{87} \mathrm{Rb}$ atoms in thermal equilibrium, write the expressions for their typical linear momentum $p$ and the typical de Broglie wavelength $\lambda_{\mathrm{dB}}$ as a function of atom's mass $m$, temperature $T$ and physical constants.

Solution: At temperature $T$, the average kinetic energy of translational motion is $\frac{3}{2} k_{\mathrm{B}} T$. Equating this result to $p^{2} /(2 m)$, we obtain typical momentum $p=\sqrt{3 m k_{\mathrm{B}} T}$ and the de Broglie wavelength

$$
\lambda_{\mathrm{dB}}=\frac{h}{p}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}
$$

Answer: $p=\sqrt{3 m k_{\mathrm{B}} T}$ , $\lambda_{\mathrm{dB}}=\frac{h}{\sqrt{3 m k_{\mathrm{B}} T}}$
Problem: C.2 Calculate the typical distance between the particles in a gas, $\ell$, as a function of particle density $n$. Hence deduce the critical temperature $T_{c}$ as a function of atom's mass, their density and physical constants.

Solution: The volume per particle $V / N$ is a good estimate for $\ell^{3}$. We obtain $\ell=n^{-1 / 3}$, with $n=N / V$ and equate $\ell=\lambda_{\mathrm{dB}}$ to express $T_{c}=h^{2} n^{2 / 3} /\left(3 m k_{\mathrm{B}}\right)$.

Answer: $\ell=n^{-1 / 3}$ , $T_{c}=\frac{h^{2} n^{2 / 3}}{3 m k_{\mathrm{B}}}$
Problem: 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: $L_{\odot}=\left(4 \pi R_{\odot}^{2}\right)\left(\sigma T_{\mathrm{s}}^{4}\right)$

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

Answer: $5.76 \times 10^{3} $
Problem: 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, $\nu=\frac{k_{\mathrm{B}} T_{\mathrm{s}}}{h} x \quad d \nu=\frac{k_{\mathrm{B}} T_{\mathrm{s}}}{h} d x$.

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

Answer: $P_{\mathrm{in}}=\frac{12 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}}$
Problem: 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: $$
\begin{aligned}
n_{\gamma}(\nu) & =\frac{u(\nu)}{h \nu} \\
& =A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right)
\end{aligned}
$$

Answer: $n_{\gamma}(\nu)=A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \nu^{2} e^ {\frac{-h \nu }{k_{\mathrm{B}} T_{\mathrm{s}}}}$
Problem: A4 Define $x_{\mathrm{g}}=h v_{\mathrm{g}} / k_{\mathrm{B}} T_{\mathrm{s}}$ where $E_{\mathrm{g}}=h v_{\mathrm{g}}$. Express the useful output power of the cell, $P_{\text {out }}$, in terms of $x_{\mathrm{g}}, A$, <br> $R_{\odot}, d_{\odot}, T_{\mathrm{s}}$ and the fundamental constants $c, h, k_{\mathrm{B}}$.

Solution: The useful power output is the useful energy quantum per photon, $E_{\mathrm{g}} \equiv h \nu_{\mathrm{g}}$, multiplied by the number of photons with energy, $E \geq E_{\mathrm{g}}$.

$$
\begin{aligned}
P_{\text {out }} & =h \nu_{\mathrm{g}} \int_{\nu_{\mathrm{g}}}^{\infty} n_{\gamma}(\nu) d \nu \\
& =h \nu_{\mathrm{g}} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}} \int_{\nu_{\mathrm{g}}}^{\infty} \nu^{2} \exp \left(-h \nu / k_{\mathrm{B}} T_{\mathrm{s}}\right) d \nu \\
& =k_{\mathrm{B}} T_{\mathrm{s}} x_{\mathrm{g}} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} \frac{2 \pi}{c^{2}}\left(\frac{k_{\mathrm{B}} T_{\mathrm{s}}}{h}\right)^{3} \int_{x_{\mathrm{g}}}^{\infty} x^{2} e^{-x} d x \\
& =\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}
\end{aligned}
$$

Answer: $P_{\text {out }}=\frac{2 \pi k_{\mathrm{B}}^{4}}{c^{2} h^{3}} T_{\mathrm{s}}^{4} A \frac{R_{\odot}^{2}}{d_{\odot}^{2}} x_{\mathrm{g}}(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2) e^{-x_{\mathrm{g}}}$
Problem: A5 Express the efficiency, $\eta$, of this solar cell in terms of $x_{\mathrm{g}}$.

Solution: Efficiency $\eta=\frac{P_{\text {out }}}{P_{\text {in }}}=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$

Answer: $\eta=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}$
Problem: A6 Make a qualitative sketch of $\eta$ versus $x_{\mathrm{g}}$. The values at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ should be clearly shown. What <br> is the slope of $\eta\left(x_{\mathrm{g}}\right)$ at $x_{\mathrm{g}}=0$ and $x_{\mathrm{g}} \rightarrow \infty$ ?

Solution: $$
\eta=\frac{1}{6}\left(x_{\mathrm{g}}^{3}+2 x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}\right) e^{-x_{\mathrm{g}}}
$$

Put limiting values, $\eta(0)=0 \quad \eta(\infty)=0$.

Since the polynomial has all positive coefficients, it increases monotonically; the exponential function decreases monotonically. Therefore, $\eta$ has only one maximum.

$$
\begin{aligned}
& \frac{\mathrm{d} \eta}{\mathrm{d} x_{\mathrm{g}}}=\frac{1}{6}\left(-x_{\mathrm{g}}^{3}+x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}} \\
& \left.\frac{\mathrm{d} \eta}{\mathrm{d} x_{\mathrm{g}}}\right|_{x_{\mathrm{g}}=0}=\left.\frac{1}{3} \quad \frac{\mathrm{d} \eta}{\mathrm{d} x_{\mathrm{g}}}\right|_{x_{\mathrm{g}} \rightarrow \infty}=0
\end{aligned}
$$

<img_4462>

Answer: $\frac{1}{3}$ , 0
Problem: A7 Let $x_{0}$ be the value of $x_{\mathrm{g}}$ for which $\eta$ is maximum. Obtain the cubic equation that gives $x_{0}$. Estimate the <br> value of $x_{0}$ within an accuracy of \pm 0.25. Hence calculate $\eta\left(x_{0}\right)$.

Solution: The maximum will be for $\frac{\mathrm{d} \eta}{\mathrm{d} x_{\mathrm{g}}}=\frac{1}{6}\left(-x_{\mathrm{g}}^{3}+x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}=0$

$$
\Rightarrow p\left(x_{\mathrm{g}}\right) \equiv x_{\mathrm{g}}^{3}-x_{\mathrm{g}}^{2}-2 x_{\mathrm{g}}-2=0
$$

A Numerical Solution by the Bisection Method:

Now,

$$
\begin{aligned}
p(0) & =-2 \\
p(1) & =-4 \\
p(2) & =-2 \\
p(3) & =10 \quad \Rightarrow \\
p(2.5) & =2.375 \quad \Rightarrow \quad 2<x_{0}<3 \\
p(2.25) & =-0.171 \quad \Rightarrow \quad 2.25<x_{0}<2.5
\end{aligned}
$$

The approximate value of $x_{\mathrm{g}}$ where $\eta$ is maximum is $x_{0}=2.27$.

Alternative methods leading to the same result are acceptable.

$\eta(2.27)=0.457$

Answer: $x_0=2.27$ , $\eta(2.27)=0.457$
Problem: A8 The band gap of pure silicon is $E_{\mathrm{g}}=1.11 \mathrm{eV}$. Calculate the efficiency, $\eta_{\mathrm{Si}}$, of a silicon solar cell using this <br> value.

Solution: $$
\begin{gathered}
x_{\mathrm{g}}=\frac{1.11 \times 1.60 \times 10^{-19}}{1.38 \times 10^{-23} \times 5763}=2.23 \\
\eta_{\mathrm{Si}}=\frac{x_{\mathrm{g}}}{6}\left(x_{\mathrm{g}}^{2}+2 x_{\mathrm{g}}+2\right) e^{-x_{\mathrm{g}}}=0.457
\end{gathered}
$$

Answer: 0.457
Problem: A9 Let us assume that the density of matter is uniform inside the Sun. Find the total gravitational potential energy, $\Omega$, of the Sun at present, in terms of $G, M_{\odot}$ and $R_{\odot}$.

Solution: The total gravitational potential energy of the Sun: $\Omega=-\int_{0}^{M_{\odot}} \frac{G m \mathrm{~d} m}{r}$

For constant density, $\rho=\frac{3 M_{\odot}}{4 \pi R_{\odot}^{3}} \quad m=\frac{4}{3} \pi r^{3} \rho \quad \mathrm{d} m=4 \pi r^{2} \rho \mathrm{d} r$

$$
\Omega=-\int_{0}^{R_{\odot}} G\left(\frac{4}{3} \pi r^{3} \rho\right)\left(4 \pi r^{2} \rho\right) \frac{\mathrm{d} r}{r}=-\frac{16 \pi^{2} G \rho^{2}}{3} \frac{R_{\odot}^{5}}{5}=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}
$$

Answer: $\Omega=-\frac{3}{5} \frac{G M_{\odot}^{2}}{R_{\odot}}$
Problem: A.10 Estimate the maximum possible time, $\tau_{\mathrm{KH}}$ (in years), for which the Sun could have been shining, according to the KH hypothesis. Assume that the luminosity of the Sun has been constant throughout this period.

Solution: $$
\begin{gathered}
\tau_{\mathrm{KH}}=\frac{-\Omega}{L_{\odot}} \\
\tau_{\mathrm{KH}}=\frac{3 G M_{\odot}^{2}}{5 R_{\odot} L_{\odot}}=1.88 \times 10^{7} \text { years }
\end{gathered}
$$

Answer: $1.88 \times 10^{7}$
Problem: B1 Calculate the flux density, $\Phi_{v}$, of the number of neutrinos arriving at the Earth, in units of $\mathrm{m}^{-2} \mathrm{~s}^{-1}$. The energy released in the above reaction is $\Delta E=4.0 \times 10^{-12} \mathrm{~J}$. Assume that the energy radiated by the Sun is entirely due to this reaction.

Solution: $$
\begin{aligned}
& \text { Solution: } \\
& \qquad \begin{array}{ll}
4.0 \times 10^{-12} \mathrm{~J} \leftrightarrow 2 \nu \\
\Rightarrow \Phi_{\nu}=\frac{L_{\odot}}{4 \pi d_{\odot}^{2} \delta E} \times 2=\frac{3.85 \times 10^{26}}{4 \pi \times\left(1.50 \times 10^{11}\right)^{2} \times 4.0 \times 10^{-12}} \times 2=6.8 \times 10^{14} \mathrm{~m}^{-2} \mathrm{~s}^{-1} .
\end{array}
\end{aligned}
$$

Answer: $6.8 \times 10^{14}$
Problem: B2 In terms of $N_{1}$ and $N_{2}$, calculate what fraction, $f$, of $v_{\mathrm{e}}$ is converted to $v_{\mathrm{x}}$.

Solution: $$
\begin{aligned}
& N_{1}=\epsilon N_{0} \\
& N_{e}=\epsilon N_{0}(1-f) \\
& N_{x}=\epsilon N_{0} f / 6 \\
& N_{2}=N_{e}+N_{x}
\end{aligned}
$$

OR

$$
\begin{gathered}
(1-f) N_{1}+\frac{f}{6} N_{1}=N_{2} \\
\Rightarrow f=\frac{6}{5}\left(1-\frac{N_{2}}{N_{1}}\right)
\end{gathered}
$$

Answer: $f=\frac{6}{5}(1-\frac{N_{2}}{N_{1}})$
Problem: B.3 Assume that an electron knocked out by a neutrino loses energy at a constant rate of $\alpha$ per unit time, while it travels through water. If this electron emits Cherenkov radiation for a time, $\Delta t$, determine the energy imparted to this electron ( $E_{\text {imparted }}$ ) by the neutrino, in terms of $\alpha, \Delta t, n, m_{\mathrm{e}}$ and $c$. (Assume the electron to be at rest before its interaction with the neutrino.)

Solution: When the electron stops emitting Cherenkov radiation, its speed has reduced to $v_{\text {stop }}=c / n$. Its total energy at this time is

$$
E_{\text {stop }}=\frac{m_{\mathrm{e}} c^{2}}{\sqrt{1-v_{\text {stop }}^{2} / c^{2}}}=\frac{n m_{\mathrm{e}} c^{2}}{\sqrt{n^{2}-1}}
$$

The energy of the electron when it was knocked out is

$$
E_{\text {start }}=\alpha \Delta t+\frac{n m_{\mathrm{e}} c^{2}}{\sqrt{n^{2}-1}}
$$

Before interacting, the energy of the electron was equal to $m_{\mathrm{e}} c^{2}$.

Thus, the energy imparted by the neutrino is

$$
E_{\text {imparted }}=E_{\text {start }}-m_{\mathrm{e}} c^{2}=\alpha \Delta t+\left(\frac{n}{\sqrt{n^{2}-1}}-1\right) m_{\mathrm{e}} c^{2}
$$

Answer: $E_{\text {imparted }}=\alpha \Delta t+(\frac{n}{\sqrt{n^{2}-1}}-1) m_{\mathrm{e}} c^{2}$