weblab-llm-competition-2025-bridge/neko-prelim-HLE_SFT_OlympiadBench huggingface dataset:
Problem: 3.3

Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ using the value of the lowest energy. The calculation should be approximated based on the following principles:

3.3.A

The potential energy of the ion should be expressed in terms of the average value of $\frac{1}{r}$.

(ie. $\frac{1}{\mathrm{r}_{\mathrm{0}}} ; \mathrm{r}_{0}$ is given in the problem).

3.3.B

In calculating the kinetic energy of the ion, use the average value of the square of the momentum given in 3.1 after being simplified by $\left(\mathrm{p}_{\mathrm{0}}\right)^{2} \cdot\left(\mathrm{r}_{\mathrm{0}}\right)^{2} \approx(\hbar)^{2}$

Solution: 3.3 

Determination of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$ :

potential energy $=-\frac{Z \cdot e^{2}}{4 \cdot \pi \cdot \varepsilon_{0} \cdot r_{0}}=-\frac{Z \cdot q^{2}}{r_{0}}$

kinetic energy $=\frac{p^{2}}{2 \cdot m}$

If the motion of the electrons is confined within the $x-y$-plane, principles of uncertainty in 3.1 can be written as

$r_{0}^{2}=(\Delta \mathrm{x})^{2}+(\Delta \mathrm{y})^{2}$

$\mathrm{p}_{0}^{2}=\left(\Delta \mathrm{p}_{\mathrm{x}}\right)^{2}+\left(\Delta \mathrm{p}_{\mathrm{y}}\right)^{2}$

$\mathrm{p}_{0}^{2}=\frac{\hbar^{2}}{4} \cdot\left[\frac{1}{(\Delta \mathrm{x})^{2}}+\frac{1}{(\Delta \mathrm{y})^{2}}\right]=\frac{\hbar^{2}}{4} \cdot\left[\frac{2}{\mathrm{r}_{0}^{2}}+\frac{2}{\mathrm{r}_{0}^{2}}\right]=\frac{\hbar^{2}}{4} \cdot \frac{4}{\mathrm{r}_{0}^{2}}$

thus

$\mathrm{p}_{0}^{2} \cdot \mathrm{r}_{0}^{2}=\hbar^{2}$

$E\left[A^{(z-1)+}\right]=\frac{p_{0}^{2}}{2 \cdot m_{e}}-\frac{Z \cdot q^{2}}{r_{0}}=\frac{\hbar^{2}}{2 \cdot m_{e} \cdot r_{e}}-\frac{Z \cdot q^{2}}{r_{0}}$

Energy minimum exists, when $\frac{\mathrm{dE}}{\mathrm{dr}_{0}}=0$.

Hence

$\frac{\mathrm{dE}}{\mathrm{dr}_{0}}=-\frac{\hbar^{2}}{\mathrm{~m}_{\mathrm{e}} \cdot \mathrm{r}_{\mathrm{e}}^{3}}+\frac{\mathrm{Z} \cdot \mathrm{q}^{2}}{\mathrm{r}_{0}^{2}}=0$

this gives $\frac{1}{r_{0}}=\frac{Z \cdot q^{2} \cdot m_{e}}{\hbar^{2}}$

hence

$$
\begin{gathered}
E\left[A^{(Z-1)+}\right]=\frac{\hbar^{2}}{2 \cdot m_{e}} \cdot\left(\frac{Z \cdot q^{2} \cdot m_{e}}{\hbar}\right)^{2}-Z \cdot q^{2} \cdot \frac{Z \cdot q^{2} \cdot m_{e}}{\hbar^{2}}=-\frac{m_{e}}{2} \cdot\left(\frac{Z \cdot q^{2}}{\hbar}\right)^{2}=-\frac{q^{2} \cdot Z^{2}}{2 \cdot r_{B}}=-E_{R} \cdot Z^{2} \\
E\left[A^{(Z-1)+}\right]=-E_{R} \cdot Z^{2}
\end{gathered}
$$

Answer: $-E_{R} \cdot Z^{2}$
Problem: 3.4

Calculate the energy of the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ taken to be in the ground state, using the same principle as the calculation of the energy of $\mathrm{A}^{(\mathrm{Z}-1)+}$. Given the average distance of each of the two electrons in the outermost shell (same as $r_{0}$ given in 3.3) denoted by $r_{1}$ and $r_{2}$, assume the average distance between the two electrons is given by $r_{1}+r_{2}$ and the average value of the square of the momentum of each electron obeys the principle of uncertainty ie.

$\mathrm{p}_{1}^{2} \cdot \mathrm{r}_{1}^{2} \approx \hbar^{2}$ and $\mathrm{p}_{2}^{2} \cdot \mathrm{r}_{2}^{2} \approx \hbar^{2}$

hint: Make use of the information that in the ground state $r_{1}=r_{2}$

Solution: 3.4 

In the case of $\mathrm{A}^{(\mathrm{Z}-1)+}$ ion captures a second electron

potential energy of both electrons $=-2 \cdot \frac{Z \cdot q^{2}}{r_{0}}$

kinetic energy of the two electrons $=2 \cdot \frac{p^{2}}{2 \cdot m}=\frac{\hbar^{2}}{m_{e} \cdot r_{0}^{2}}$

potential energy due to interaction between the two electrons $=\frac{q^{2}}{\left|\vec{r}_{1}-\vec{r}_{2}\right|}=\frac{q^{2}}{2 \cdot r_{0}}$

$E\left[A^{(z-2)+}\right]=\frac{\hbar^{2}}{m_{e} \cdot r_{0}^{2}}-\frac{2 \cdot Z \cdot q^{2}}{r_{0}^{2}}+\frac{q^{2}}{2 \cdot r_{0}}$

total energy is lowest when $\frac{\mathrm{dE}}{\mathrm{dr}}=0$

hence

$0=-\frac{2 \cdot \hbar^{2}}{m_{e} \cdot r_{0}^{3}}+\frac{2 \cdot Z \cdot q^{2}}{r_{0}^{3}}-\frac{q^{2}}{2 \cdot r_{0}^{2}}$

hence

$\frac{1}{r_{0}}=\frac{q^{2} \cdot m_{e}}{2 \cdot \hbar^{2}} \cdot\left(2 \cdot z-\frac{1}{2}\right)=\frac{1}{r_{B}} \cdot\left(z-\frac{1}{4}\right)$

$E\left[A^{(Z-2)+}\right]=\frac{\hbar^{2}}{m_{e}} \cdot\left(\frac{q^{2} \cdot m_{e}}{2 \cdot \hbar^{2}}\right)^{2}-\frac{q^{2} \cdot\left(2 \cdot Z-\frac{1}{2}\right)}{\hbar} \cdot \frac{q^{2} \cdot m_{e} \cdot\left(2 \cdot Z-\frac{1}{2}\right)}{2 \cdot \hbar}$

$E\left[A^{(Z-2)+}\right]=-\frac{m_{e}}{4} \cdot\left[\frac{q^{2} \cdot\left(2 \cdot Z-\frac{1}{2}\right)}{\hbar}\right]^{2}=-\frac{m_{e} \cdot\left[q^{2} \cdot\left(Z-\frac{1}{4}\right)\right]^{2}}{\hbar^{2}}=-\frac{q^{2} \cdot\left(Z-\frac{1}{4}\right)^{2}}{\hbar^{2}}$

this gives

$$
E\left[A^{(z-2)+}\right]=-2 \cdot E_{R} \cdot\left(z-\frac{1}{4}\right)^{2}
$$

Answer: $-2 \cdot E_{R} \cdot(z-\frac{1}{4})^{2}$
Problem: 3.5

Consider in particular the ion $\mathrm{A}^{(\mathrm{Z}-2)+}$ is at rest in the ground state when capturing an additional electron and the captured electron is also at rest prior to the capturing. Determine the numerical value of $Z$, if the frequency of the emitted photon accompanying electron capturing is $2,057\times 10^{17} \mathrm{rad} / \mathrm{s}$. Identify the element which gives rise to the ion.

Solution: 3.5 

The ion $\mathrm{A}^{(\mathrm{Z}-1)+}$ is at rest when it captures the second electron also at rest before capturing.

From the information provided in the problem, the frequency of the photon emitted is given by

$v=\frac{\omega}{2 \cdot \pi}=\frac{2,057 \cdot 10^{17}}{2 \cdot \pi} \mathrm{Hz}$

The energy equation can be simplified to $\quad E\left[A^{(Z-1)+}\right]-E\left[A^{(Z-2)+}\right]=\hbar \cdot \omega=h \cdot v$ that is

$-E_{R} \cdot Z^{2}-\left[-2 \cdot E_{R} \cdot\left(Z-\frac{1}{4}\right)^{2}\right]=\hbar \cdot \omega$

putting in known numbers follows

$2,180 \cdot 10^{-18} \cdot\left[-Z^{2}+2 \cdot\left(Z-\frac{1}{4}\right)^{2}\right]=1,05 \cdot 10^{-34} \cdot 2,607 \cdot 10^{17}$

this gives

$Z^{2}-Z-12,7=0$

with the physical sensuous result $\quad Z=\frac{1+\sqrt{1+51}}{2}=4,1$

This implies $\mathrm{Z}=4$, and that means Beryllium



Answer: Z=4, Beryllium
Problem: A. Determine the boiling temperatures of the liquids A and B under the pressure $p_{0}$.

Solution: PART A

The liquid boils when the pressure of its saturated vapor is equal to the external pressure. Thus, in order to find the boiling temperature of the liquid $i(i-\mathrm{A}$ or $\mathrm{B}$ ), one should determine such a temperature $T_{b i}$ (or $t_{b i}$ ) for which $p_{i} / p_{0}=1$.

Then $\ln \left(p_{i} / p_{0}\right)=0$, and we have:

$$
T_{b i}=-\frac{\alpha_{i}}{\beta_{i}}
$$



The coefficients $\alpha_{i}$ and $\beta_{i}$ are not given explicitly. However, they can be calculated from the formula given in the text of the problem. For this purpose one should make use of the numerical data given in the Tab. 1.1.

For the liquid A, we have:

$$
\begin{aligned}
& \ln 0.284=\frac{\alpha_{A}}{(40+273.15) \mathrm{K}}+\beta_{A}, \\
& \ln 1.476=\frac{\alpha_{A}}{(90+273.15) \mathrm{K}}+\beta_{A} .
\end{aligned}
$$

After subtraction of these equations, we get:

$$
\begin{gathered}
\ln 0.284-\ln 1.476=\alpha_{A}\left(\frac{1}{40+273.15}-\frac{1}{90+273.15}\right) \mathrm{K}^{-1} . \\
\alpha_{A}=\frac{\ln \frac{0.284}{1.476}}{\frac{1}{40+273.15}-\frac{1}{90+273.15}} \mathrm{~K} \approx-3748.49 \mathrm{~K}
\end{gathered}
$$

Hence,

$$
\beta_{A}=\ln 0.284-\frac{\alpha_{A}}{(40+273.15) \mathrm{K}} \approx 10.711
$$

Thus, the boiling temperature of the liquid $\mathrm{A}$ is equal to

$$
T_{b A}=3748.49 \mathrm{~K} / 10.711 \approx 349.95 \mathrm{~K} .
$$

In the Celsius scale the boiling temperature of the liquid $\mathrm{A}$ is

$$
t_{b A}=(349.95-273.15)^{\circ} \mathrm{C}=76.80^{\circ} \mathrm{C} \approx 77^{\circ} \mathrm{C} .
$$

For the liquid B, in the same way, we obtain:

$$
\begin{aligned}
\alpha_{B} & \approx-5121.64 \mathrm{~K} \\
\beta_{B} & \approx 13.735 \\
T_{b B} & \approx 372-89 \mathrm{~K} \\
t_{b B} & \approx 99.74^{\circ} \mathrm{C} \approx 100^{\circ} \mathrm{C} .
\end{aligned}
$$

Answer: $76.8$ , $100$
Problem: B. How many times would the resolving power of the above microscope increase or decrease if the electron beam were replaced with the proton beam? Assume that the resolving power of the microscope (i.e. the smallest distance between two point objects whose circular images can be just separated) depends only on the wave properties of the particles.

Assume that the velocities of the electrons and protons before their acceleration are zero, and that there is no interaction between own magnetic moment of either electrons or protons and the magnetic field. Assume also that the electromagnetic radiation emitted by the moving particles can be neglected.

NOTE: Very often physicists use 1 electron-volt $(1 \mathrm{eV})$, and its derivatives such as 1 $\mathrm{keV}$ or $1 \mathrm{MeV}$, as a unit of energy. 1 electron-volt is the energy gained by the electron that passed the potential difference equal to $1 \mathrm{~V}$.

Perform the calculations assuming the following data:

$$
\begin{array}{ll}
\text { Rest energy of electron: } & E_{e}=m_{e} c^{2}=511 \mathrm{keV} \\
\text { Rest energy of proton: } & E_{p}=m_{p} c^{2}=938 \mathrm{MeV}
\end{array}
$$

Solution: PART B

The resolving power of the microscope (in the meaning mentioned in the text of the problem) is proportional to the wavelength, in our case to the length of the de Broglie wave:

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

where $h$ denotes the Planck constant and $p$ is the momentum of the particle. We see that $\lambda$ is inversely proportional to the momentum of the particle. Therefore, after replacing the electron beam with the proton beam the resolving power will be changed by the factor $p / p^{\prime} \approx 1 / 35$. It means that our proton microscope would allow observation of the objects about 35 times smaller than the electron microscope.

Answer: 35
Problem: 1.1. Assume that the temperature of the atmosphere is uniform and equal to $T_{0}$. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $z$.

Solution: For an altitude change $d z$, the atmospheric pressure change is :

$$
d p=-\rho g d z
\tag{1}
$$

where $g$ is the acceleration of gravity, considered constant, $\rho$ is the specific mass of air, which is considered as an ideal gas:

$$
\rho=\frac{m}{V}=\frac{p \mu}{R T}
$$

Put this expression in (1) :

$$
\frac{d p}{p}=-\frac{\mu g}{R T} d z
$$

If the air temperature is uniform and equals $T_{0}$, then

$$
\frac{d p}{p}=-\frac{\mu g}{R T_{0}} d z
$$

After integration, we have :

$$
p(z)=p(0) \mathbf{e}^{-\frac{\mu g}{R T_{0}} z}
\tag{2}
$$

Answer: $p(z)=p(0) e^{-\frac{\mu g}{R T_{0}} z}$
Problem: 1.2.1. Write down the expression giving the atmospheric pressure $p$ as a function of the altitude $Z$.

Solution: If

$$
T(z)=T(0)-\Lambda z
\tag{3}
$$

then

$$
\frac{d p}{p}=-\frac{\mu g}{R[T(0)-\Lambda z]} d z
\tag{4}
$$

Knowing that :

$$
\int \frac{d z}{T(0)-\Lambda z}=-\frac{1}{\Lambda} \int \frac{d[T(0)-\Lambda z]}{T(0)-\Lambda z}=-\frac{1}{\Lambda} \ln (T(0)-\Lambda z)
$$

by integrating both members of (4), we obtain :

$$
\ln \frac{p(z)}{p(0)}=\frac{\mu g}{R \Lambda} \ln \frac{T(0)-\Lambda z}{T(0)}=\frac{\mu g}{R \Lambda} \ln \left(1-\frac{\Lambda z}{T(0)}\right)
$$
$$
p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}
\tag{5}
$$

Answer: $p(z)=p(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}}$
Problem: 1.2.2. A process called free convection occurs when the air density increases with altitude. At which values of $\Lambda$ does the free convection occur?

Solution: The free convection occurs if:

$$
\frac{\rho(z)}{\rho(0)}>1
$$

The ratio of specific masses can be expressed as follows:

$$
\frac{\rho(z)}{\rho(0)}=\frac{p(z)}{p(0)} \frac{T(0)}{T(z)}=\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\mu g}{R \Lambda}-1}
$$

The last term is larger than unity if its exponent is negative:

$$
\frac{\mu g}{R \Lambda}-1<0
$$

Then :

$$
\Lambda>\frac{\mu g}{R}=\frac{0.029 \times 9.81}{8.31}=0.034 \frac{\mathrm{K}}{\mathrm{m}}
$$

Answer: $0.034$
Problem: 2.1. The change of the parcel temperature $T_{\text {parcel }}$ with altitude is defined by $\frac{d T_{\text {parcel }}}{d z}=-G$. Derive the expression of $G\left(T, T_{\text {parcel }}\right)$.

Solution: In vertical motion, the pressure of the parcel always equals that of the surrounding air, the latter depends on the altitude. The parcel temperature $T_{\text {parcel }}$ depends on the pressure.

We can write:

$$
\frac{d T_{\text {parcel }}}{d z}=\frac{d T_{\text {parcel }}}{d p} \frac{d p}{d z}
$$

$p$ is simultaneously the pressure of air in the parcel and that of the surrounding air.

Expression for $\frac{d T_{\text {parcel }}}{d p}$

By using the equation for adiabatic processes $p V^{\gamma}=$ const and equation of state, we can deduce the equation giving the change of pressure and temperature in a quasi-equilibrium adiabatic process of an air parcel:

$$
T_{\text {parcel }} p^{\frac{1-\gamma}{\gamma}}=\text { const }
\tag{6}
$$



where $\gamma=\frac{c_{p}}{c_{V}}$ is the ratio of isobaric and isochoric thermal capacities of air. By logarithmic differentiation of the two members of (6), we have:

$$
\frac{d T_{\text {parcel }}}{T_{\text {parcel }}}+\frac{1-\gamma}{\gamma} \frac{d p}{p}=0
$$

Or

$$
\frac{d T_{\text {parcel }}}{d p}=\frac{T_{\text {parcel }}}{p} \frac{\gamma-1}{\gamma}
\tag{7}
$$

Note: we can use the first law of thermodynamic to calculate the heat received by the parcel in an elementary process: $d Q=\frac{m}{\mu} c_{V} d T_{\text {parcel }}+p d V$, this heat equals zero in an adiabatic process. Furthermore, using the equation of state for air in the parcel $p V=\frac{m}{\mu} R T_{\text {parcel }}$ we can derive (6)

Expression for $\frac{d p}{d z}$

From (1) we can deduce:

$$
\frac{d p}{d z}=-\rho g=-\frac{p g \mu}{R T}
$$

where $T$ is the temperature of the surrounding air.

On the basis of these two expressions, we derive the expression for $d T_{\text {parcel }} / d z$ :

$$
\frac{d T_{\text {parcel }}}{d z}=-\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}=-G
\tag{8}
$$

In general, $G$ is not a constant.

Answer: $\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{T_{\text {parcel }}}{T}$
Problem: 2.2.1. Derive the expression of $\Gamma$

Solution: If at any altitude, $T=T_{\text {parcel }}$, then instead of $G$ in (8), we have :

$$
\Gamma=\frac{\gamma-1}{\gamma} \frac{\mu g}{R}=\mathrm{const}
\tag{9}
$$

or



$$
\Gamma=\frac{\mu g}{c_{p}}
\tag{9'}
$$

Answer: $\Gamma=\frac{\mu g}{c_{p}}$
Problem: 2.2.2. Calculate the numerical value of $\Gamma$.

Solution: Numerical value:

$$
\Gamma=\frac{1.4-1}{1.4} \frac{0.029 \times 9.81}{8.31}=0.00978 \frac{\mathrm{K}}{\mathrm{m}} \approx 10^{-2} \frac{\mathrm{K}}{\mathrm{m}}
$$

Answer: $0.00978 $
Problem: 2.2.3. Derive the expression of the atmospheric temperature $T(z)$ as a function of the altitude.

Solution: Thus, the expression for the temperature at the altitude $z$ in this special atmosphere (called adiabatic atmosphere) is :

$$
T(z)=T(0)-\Gamma z
\tag{10}
$$

Answer: $T(z)=T(0)-\Gamma z$
Problem: 2.3. Assume that the atmospheric temperature depends on altitude according to the relation $T(z)=T(0)-\Lambda z$, where $\Lambda$ is a constant. Find the dependence of the parcel temperature $T_{\text {parcel }}(z)$ on altitude $z$.

Solution: Search for the expression of $T_{\text {parcel }}(z)$

Substitute $T$ in (7) by its expression given in (3), we have:

$$
\frac{d T_{\text {parcel }}}{T_{\text {parcel }}}=-\frac{\gamma-1}{\gamma} \frac{\mu g}{R} \frac{d z}{T(0)-\Lambda z}
$$

Integration gives:

$$
\ln \frac{T_{\text {parcel }}(z)}{T_{\text {parcel }}(0)}=-\frac{\gamma-1}{\gamma} \frac{\mu g}{R}\left(-\frac{1}{\Lambda}\right) \ln \frac{T(0)-\Lambda z}{T(0)}
$$

Finally, we obtain:

$$
T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}
\tag{11}
$$

Answer: $T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(\frac{T(0)-\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}$
Problem: 2.4. Write down the approximate expression of $T_{\text {parcel }}(z)$ when $|\Lambda z|<<T(0)$ and $T(0) \approx T_{\text {parcel }}(0)$.

Solution: From (11) we obtain

$$
T_{\text {parcel }}(z)=T_{\text {parcel }}(0)\left(1-\frac{\Lambda z}{T(0)}\right)^{\frac{\Gamma}{\Lambda}}
$$

If $\Lambda z<<T(0)$, then by putting $x=\frac{-T(0)}{\Lambda z}$, we obtain

$$
\begin{aligned}
T_{\text {parcel }}(z) & =T_{\text {parcel }}(0)\left(\left(1+\frac{1}{x}\right)^{x}\right)^{-\frac{\Gamma z}{T(0)}} \\
& \approx T_{\text {parcel }}(0) \mathbf{e}^{-\frac{\Gamma z}{T(0)}} \approx T_{\text {parcel }}(0)\left(1-\frac{\Gamma z}{T(0)}\right) \approx T_{\text {parcel }}(0)-\Gamma z
\end{aligned}
$$



hence,

$$
T_{\text {parcel }}(z) \approx T_{\text {parcel }}(0)-\Gamma z
\tag{12}
$$

Answer: $T_{\text {parcel }}(0)-\Gamma z$
Problem: 3.2. A parcel has its temperature on ground $T_{\text {parcel }}(0)$ higher than the temperature $T(0)$ of the surrounding air. The buoyancy force will make the parcel rise. Derive the expression for the maximal altitude the parcel can reach in the case of a stable atmosphere in terms of $\Lambda$ and $\Gamma$.

Solution: In a stable atmosphere, with $\Lambda<\Gamma$, a parcel, which on ground has temperature $T_{\text {parcel }}(0)>T(0)$ and pressure $p(0)$ equal to that of the atmosphere, can rise and reach a maximal altitude $h$, where $T_{\text {parcel }}(h)=T(h)$.

In vertical motion from the ground to the altitude $h$, the air parcel realizes an adiabatic quasi-static process, in which its temperature changes from $T_{\text {parcel }}(0)$ to $T_{\text {parcel }}(h)=T(h)$. Using (11), we can write:

$$
\begin{aligned}
& \left(1-\frac{\Lambda h}{T(0)}\right)^{-\frac{\Gamma}{\Lambda}}=\frac{T_{\text {parcel }}(0)}{T(h)}=\frac{T_{\text {parcel }}(0)}{T(0)\left(1-\frac{\Lambda h}{T(0)}\right)} \\
& \left(1-\frac{\Lambda h}{T(0)}\right)^{1-\frac{\Gamma}{\Lambda}}=T_{\text {parcel }}(0) \times T^{-1}(0) \\
& 1-\frac{\Lambda h}{T(0)}=T_{\text {parcel }}^{\frac{\Lambda}{\Lambda-\Gamma}}(0) \times T^{-\frac{\Lambda}{\Lambda-\Gamma}}(0) \\
& h=\frac{1}{\Lambda} T(0)\left[1-T_{\text {parcel }}^{\frac{\Lambda}{\Lambda-\Gamma}}(0) \times T^{-\frac{\Lambda}{\Lambda-\Gamma}}(0)\right] \\
& =\frac{1}{\Lambda}\left[T(0)-T_{\text {parcel }}^{-\frac{\Lambda}{\Lambda-\Gamma}}(0) T^{\frac{\Gamma}{\Gamma-\Lambda}}(0)\right]
\end{aligned}
$$

So that the maximal altitude $h$ has the following expression:

$$
h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]
\tag{13}
$$

Answer: $h=\frac{1}{\Lambda}\left[T(0)-\left(\frac{(T(0))^{\Gamma}}{\left(T_{\text {parcel }}(0)\right)^{\Lambda}}\right)^{\frac{1}{\Gamma-\Lambda}}\right]$
Problem: 4.1. Table 1 shows air temperatures recorded by a radio sounding balloon at 7:00 am on a November day in Hanoi. The change of temperature with altitude can be approximately described by the formula $T(z)=T(0)-\Lambda z$ with different lapse rates $\Lambda$ in the three layers $0<\mathrm{z}<96 \mathrm{~m}, 96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}$ and $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}$.

Consider an air parcel with temperature $T_{\text {parcel }}(0)=22^{\circ} \mathrm{C}$ ascending from ground. On the basis of the data given in Table 1 and using the above linear approximation, calculate the temperature of the parcel at the altitudes of $96 \mathrm{~m}$ and $119 \mathrm{~m}$.

Solution: Using data from the Table, we obtain the plot of $z$ versus $T$ shown in Figure 3.

<img_4370>

Figure 3

We can divide the atmosphere under 200m into three layers, corresponding to the following altitudes:

1) $0<z<96 \mathrm{~m}, \quad \Lambda_{1}=\frac{21.5-20.1}{91}=15.4 \times 10^{-3} \frac{\mathrm{K}}{\mathrm{m}}$.
2) $96 \mathrm{~m}<\mathrm{z}<119 \mathrm{~m}, \Lambda_{2}=0$, isothermal layer.
3) $119 \mathrm{~m}<\mathrm{z}<215 \mathrm{~m}, \Lambda_{3}=-\frac{22-20.1}{215-119}=-0.02 \frac{\mathrm{K}}{\mathrm{m}}$.

In the layer 1), the parcel temperature can be calculated by using (11)

$$
T_{\text {parcel }}(96 \mathrm{~m})=294.04 \mathrm{~K} \approx 294.0 \mathrm{~K} \text { that is } 21.0^{\circ} \mathrm{C}
$$

In the layer 2), the parcel temperature can be calculated by using its expression in isothermal atmosphere $T_{\text {parcel }}(z)=T_{\text {parcel }}(0) \exp \left[-\frac{\Gamma z}{T(0)}\right]$.



The altitude $96 \mathrm{~m}$ is used as origin, corresponding to $0 \mathrm{~m}$. The altitude $119 \mathrm{~m}$ corresponds to $23 \mathrm{~m}$. We obtain the following value for parcel temperature:

$$
T_{\text {parcel }}(119 \mathrm{~m})=293.81 \mathrm{~K} \text { that is } 20.8^{\circ} \mathrm{C}
$$

Answer: $294.04$, $293.81 $
Problem: 4.2. Determine the maximal elevation $H$ the parcel can reach, and the temperature $T_{\text {parcel }}(H)$ of the parcel.

$H$ is called the mixing height. Air pollutants emitted from ground can mix with the air in the atmosphere (e.g. by wind, turbulence and dispersion) and become diluted within this layer.

Solution: In the layer 3), starting from $119 \mathrm{~m}$, by using (13) we find the maximal elevation $h=23 \mathrm{~m}$, and the corresponding temperature $293.6 \mathrm{~K}$ (or $20.6{ }^{\circ} \mathrm{C}$ ).

Finally, the mixing height is

$$
H=119+23=142 \mathrm{~m} .
$$

And

$$
T_{\text {parcel }}(142 \mathrm{~m})=293.6 \mathrm{~K} \text { that is } 20.6^{\circ} \mathrm{C}
$$

From this relation, we can find $T_{\text {parcel }}(119 \mathrm{~m}) \approx 293.82 \mathrm{~K}$ and $h=23 \mathrm{~m}$.

Note: By using approximate expression (12) we can easily find $T_{\text {parcel }}(z)=294 \mathrm{~K}$ and 293.8 K at elevations $96 \mathrm{~m}$ and $119 \mathrm{~m}$, respectively. At $119 \mathrm{~m}$ elevation, the difference between parcel and surrounding air temperatures is $0.7 \mathrm{~K}$ (= 293.8 - 293.1), so that the maximal distance the parcel will travel in the third layer is $0.7 /\left(\Gamma-\Lambda_{3}\right)=0.7 / 0.03=23 \mathrm{~m}$.

Answer: 142, 293.6
Problem: A.1 Consider a system of two stars with masses $M_{1}, M_{2}$, at locations $\vec{r}_{1}, \vec{r}_{2}$, respectively, with respect to the center-of-mass of the system, that is,

$$
M_{1} \overrightarrow{r_{1}}+M_{2} \overrightarrow{r_{2}}=0 .
\tag{!}
$$

The stars are isolated from the rest of the Universe and moving at nonrelativistic velocities. Using Newton's laws, the acceleration vector of mass $M_{1}$ can be expressed as

$$
\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\alpha \frac{\vec{r}_{1}}{r_{1}^{n}}
\tag{2}
$$

where $r_{1}=\left|\vec{r}_{1}\right|, r_{2}=\left|\vec{r}_{2}\right|$. Find $n \in \mathbb{N}$ and $\alpha=\alpha\left(G, M_{1}, M_{2}\right)$, where $G$ is Newton's constant $\left[G \simeq 6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^{2} \mathrm{~kg}^{-2}\right]$.

Solution: Apply Newton's law to mass $M_{1}$ :

$$
M_{1} \frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=G \frac{M_{1} M_{2}}{\left|\vec{r}_{2}-\vec{r}_{1}\right|^{2}} \frac{\vec{r}_{2}-\vec{r}_{1}}{\left|\vec{r}_{2}-\overrightarrow{r_{1}}\right|}
\tag{1}
$$

Use, from eq. (1) of the question sheet

$$
\vec{r}_{2}=-\frac{M_{1}}{M_{2}} \vec{r}_{1}
\tag{2}
$$

in eq. (1) above, to obtain

$$
\frac{\mathrm{d}^{2} \vec{r}_{1}}{\mathrm{~d} t^{2}}=-\frac{G M_{2}^{3}}{\left(M_{1}+M_{2}\right)^{2} r_{1}^{2}} \frac{\vec{r}_{1}}{r_{1}} .
\tag{3}
$$

Answer: $n=3$ , $\alpha=\frac{G M_{2}^{3}}{(M_{1}+M_{2})^{2}}$