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IZhO 2013, Теория, английски — Задача 3. Bohr’s model of a hydrogen atom

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Всички материали от IZhO 2013

IX International Zhautykov Olympiad/Theoretical Competition · 15 януари 2013 г. · 10 т.

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The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
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Begin each question on a separate sheet of paper.
Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.
At the end of the exam, arrange all sheets for each problem in the following order:
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Условие

Consider a hydrogen atom, which consists of a proton, whose mass can be assumed virtually infinite, and an electron of mass $m_e=9.11\times10^{-31}\ kg$ . The proton has a positive charge of $e=1.61\times10^{-19}$ Cl and the electron has a negative charge $-e$ , so that the atom is totally neutral.

1. It is known that most of the atoms are stable and can exist for a long time. This means that the electron being in the hydrogen atom cannot be at rest since the attraction force would inevitably make it fall onto the proton. Let the distance between the electron and the proton be equal to $r_0=5.00\times10^{-11}\ m/$ Assuming that the laws of classical physics are applicable, find the time $t_1$ that the initially motionless electron needs to fall onto the proton. [2 т.]

Hence, to guarantee the stability of the hydrogen atom, the electron must move around the proton like the planets revolve around the Sun. Let the electron orbit be circular and assume again that the laws of classical physics can still be applied.

2. Find the electron velocity $v$ as a function of the orbit radius $r$ . [1 т.]

3. Find the angular momentum $L$ of the electron as a function of the orbit radius $r$ . [0,5 т.]

Danish physicist Niels Bohr postulated that the angular momentum of the electron is an integer of the Planck constant $\hbar=1.05\times10^{-34}$ Дж∙с, that is $L=n\hbar$ , where $n$ is the main quantum number.

4. Find the possible radii $r_n$ of the electron orbits in the hydrogen atom. [0,5 т.]

5. Calculate the numerical value of the minimum radius $r_n$ of the electron orbit in the hydrogen atom. [0,5 т.]

6. Find the possible values of the total energy $E_n$ of the electron in the hydrogen atom. [1 т.]

7. Calculate the numerical value of the minimum total energy $E_1$ of the electron in the hydrogen atom. [0,5 т.]

According to classical electrodynamics, if the electron is accelerated, it loses its energy due to electromagnetic radiation. The power of the radiation is given by

P=\frac{1}{6\pi\varepsilon_0}\frac{e^2a^2}{c^3},

where $a$ is the electron acceleration, $\varepsilon_0=8.85\times10^{-12}\ F/m$ is the dielectric constant, $c=3.00\times10^8\ m/s$ is the speed of light in vacuum.
Electromagnetic radiation would lead to the fact that in the Bohr model the electron would fall onto the proton. In the following assume that the electron orbit is almost circular.

8. Assuming that at the time moment $t=0$ the electron moves in the orbit with the radius $r_1$ , find the dependence of the orbit radius on time $t$ . [1,5 т.]

9. Find and calculate the time $\tau_1$ that is needed for the electron to fall from the orbit with the radius $r_1$ onto the proton. [0,5 т.]

10. How many revolutions around the proton can the electron make during its fall for the time interval $\tau_1$ ? [2 т.]

Отговори

Решение

Bohr model for hydrogen atom

We can use Kepler’s third law to find the free fall time of the electron onto the proton. Consider a circular orbit of radius $R$ , then the equation of motion of the electron can be written as

m_e\left(\frac{2\pi}{T}\right)^2R=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{R^2}, \qquad (1)

where $T$ is the period of revolution.
Thus, Kepler's third law for the electron is given by

\frac{a^3}{T^2}=\frac{1}{16\pi^3\varepsilon_0}\frac{e^2}{m_e}, \qquad (2)

where $a$ is the major semi-axis of an elliptic orbit.
Consider the free fall of the electron onto the proton as a motion along a very elongated ellipse with semi-axes $a=r_0/2$ . Then, the free fall time of the electron equals to the half of this period

t_1=\frac{T}{2}=\sqrt{\frac{\pi^3m_e\varepsilon_0r_0^3}{2e^2}}=2.46\times10^{-17}\ \text{с}. \qquad (3)

To find the dependence of the electron velocity on the orbit radius we again use Newton's second law, which is now written in the form

m_e\frac{v^2}{r}=\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2}, \qquad (4)

which yields

v=\sqrt{\frac{1}{4\pi\varepsilon_0}\frac{e^2}{m_er}}, \qquad (5)

The angular momentum of the electron is found from equation (5) as

L=m_evr=\sqrt{\frac{m_ere^2}{4\pi\varepsilon_0}}. \qquad (6)

Since, according to the Bohr model of the hydrogen atom the angular momentum of the electron is quantized, that is $L=n\hbar$ , From equation (6) we can get

r_n=\frac{4\pi\varepsilon_0n^2\hbar^2}{m_ee^2}. \qquad (7)

From equation (7) we see that the minimal radius corresponds to the quantum number $n=1$ , hence

r_1=\frac{4\pi\varepsilon_0\hbar^2}{m_ee^2}=5.19\times10^{-11}\ \text{м}. \qquad (8)

The total energy of the electron in the atom is the sum of the kinetic and potential energies. Taking into account equation (5) we can write the total energy of electron in the following form

E=\frac{m_ev^2}{2}-\frac{e^2}{4\pi\varepsilon_0r}=-\frac{e^2}{8\pi\varepsilon_0r}. \qquad (9)

Substituting the possible values of the orbit radii of the electron (7), we immediately obtain

E_n=-\frac{m_ee^4}{32\pi^2\varepsilon_0^2\hbar^2n^2}. \qquad (10)

From equation (10) we see that the minimal total energy corresponds to the quantum number $n=1$ , hence

E_1=-\frac{m_ee^4}{32\pi^2\varepsilon_0^2\hbar^2}=-2.24\times10^{-18}\ \text{Дж}. \qquad (11)

The total energy of the electron (9) is spent on the emission of electromagnetic waves

\frac{dE}{dt}=-P. \qquad (12)

From the equation (9) and (12) we get

r(t)^2\frac{dr(t)}{dt}=-\frac{e^4}{12\pi^2\varepsilon_0^2m_e^2c^3}. \qquad (13)

By solving equation (13) with the initial condition $r(0)=r_1$ , we get

r(t)=\sqrt[3]{r_1^3-\frac{e^4}{4\pi^2\varepsilon_0^2m_e^2c^3}t}. \qquad (14)

The falling time $\tau_1$ is found from the condition $r(t)=0$ . By substituting this into equation (14)

\tau_1=\frac{4\pi^2m_e^2\varepsilon_0^2c^3r_1^3}{e^4}=\frac{256\pi^5\varepsilon_0^5c^3\hbar^5}{m_ee^{10}}=1.44\times10^{-11}\ \text{с}. \qquad (15)

|For a short time interval $dt$ the electron makes a turn by the angle $d\varphi$ , defined as

d\varphi=\frac{v}{r}dt. \qquad (16)

By substituting the velocity from equation (5) and $dt$ from equation (13), we find

d\varphi=-\frac{6c^3(\pi m_e\varepsilon_0)^{3/2}}{e^3}\sqrt{r}dr. \qquad (17)

Total angle of rotation is found by the integration from $r_1$ to zero

\varphi=\int_{r_1}^{0}d\varphi=\int_0^{r_1}\frac{6c^3(\pi m_e\varepsilon_0)^{3/2}}{e^3}\sqrt{r}dr=\frac{4c^3(\pi m_e\varepsilon_0r_1)^{3/2}}{e^3}=\frac{32\pi^3\varepsilon_0^3c^3\hbar^3}{e^6}. \qquad (18)

Thus, the total number of revolutions is equal to

N=\frac{\varphi}{2\pi}=\frac{16\pi^2\varepsilon_0^3c^3\hbar^3}{e^6}=1.96\times10^5. \qquad (19)

Marking scheme

№	Description	points
1	Eq (1)	0.5	2.0
	Eq(2)	0.5
	Eq(3)	0.5
	Correct numerical value of $t_1$	0.5
2	Eq (4)	0.5	1,0
	Eq (5)	0.5
3	Eq (6)	0.5	0.5
4	Eq (7)	0.5	0.5
5	Correct numerical value of $r_1$	0.5	0.5
6	Eq (9)	0.5	1.0
	Eq (10)	0.5
7	Correct numerical value of $E_1$	0.5	0.5
8	The conservation law (12)	0.5	1.5
	Eq (13)	0.5
	Eq (14)	0.5
9	Correct numerical value $\tau_1$	0.5	0.5
10	Eq (16)	0.5	2.0
	Eq(17)	0.5
	Eq(18)	0.5
	Correct numerical value $N$	0.5
Total			10,0

Оригинал в Архива: IZhO_2013_Theory_eng_fiz.doc · официални решения: IZhO_2013_Theory_eng_sol_fiz.doc