Съдържание

INPhO 2021, Теория — Задача 3

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от INPhO 2021

Indian Olympiad Qualifier in Physics (IOQP) 2020-2021 — Part II: Indian National Physics Olympiad (INPhO) · 7 февруари 2021 г.

Указания

This booklet consists of 4 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheets in the Answer Booklet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations.
Strike out any rough work that you do not want to be considered for evaluation. You may also use the space on the Question Paper for rough work – this will NOT be evaluated.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Last page of the question paper can be used for rough work.
Please submit the Answer Booklet at the end of the examination. You may retain the Question Paper.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00 \times 10^{8}\ \mathrm{m{\cdot}s^{-1}}$
Planck's constant	$h$	$6.63 \times 10^{-34}\ \mathrm{J{\cdot}s}$
Magnitude of electron charge	$e$	$1.60 \times 10^{-19}\ \mathrm{C}$
Rest mass of electron	$m_e$	$9.11 \times 10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00 \times 10^{9}\ \mathrm{N{\cdot}m^{2}{\cdot}C^{-2}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m{\cdot}s^{-2}}$

Note on solutions

Please note that alternate/equivalent methods and different way of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

Consider an electron (mass $m$ , magnitude of charge $e$ ) moving initially around a nucleus of charge $2e$ in a circular orbit of radius $10^{-10}\,\mathrm{m}$ . In this problem we use SI units throughout and neglect all relativistic effects.

(a) Obtain the expression for the frequency, $f$ , of the electron in the circular orbit (numerical value is not required). [2 т.]

From classical electrodynamics, we know that an accelerated electron radiates energy. The expression for the power $P$ of this radiation is given by

P = K\epsilon_0^{w} e^{x} a^{y} c^{z}

where $a$ is the acceleration, $c$ is the speed of light, $\epsilon_0$ is the permittivity of free space, and $K$ is a dimensionless constant.

(b) Obtain $\{w, x, y, z\}$ using dimensional analysis. [2 т.]

Due to the loss of energy through radiation, the electron does not remain in the circular orbit, and gradually spirals into the nucleus. Take the constant $K$ to be $5.31 \times 10^{-2}$ .

(c) Let $T$ be the time it takes for the electron to reach the nucleus. Calculate $T$ if the radius of the nucleus is $10^{-14}\,\mathrm{m}$ . [5 т.]

Отговори

Решение

(a) Solution:

The centripetal force for the circular motion of the electron is provided by the Coulomb attraction of the nucleus. Let $r$ be the radius of the circular orbit, and $v$ the speed of the electron in this orbit, then

\frac{mv^2}{r} = \frac{2e^2}{4\pi\epsilon_0 r^2} \qquad (3.1)

f = \frac{v}{2\pi r} = \left(\frac{2}{4\pi\epsilon_0 m}\right)^{1/2} \frac{e}{2\pi r^{3/2}} \qquad (3.2)

(b) Solution: $w = -1, x = 2, y = 2, z = -3$

(c) Solution:

The total energy of an electron in the orbit is

E(r) = -\frac{1}{4\pi\epsilon_0}\frac{e^2}{r} \qquad (3.3)

-\dot{E}(r) = -\frac{1}{4\pi\epsilon_0}\frac{e^2\dot{r}}{r^2} \qquad (3.4)

the acceleration is

a = \frac{v^2}{r} = \frac{1}{4\pi\epsilon_0}\frac{2e^2}{mr^2} \qquad (3.5)

We use Eq. (3.5) in the power radiated, which yields the energy loss rate

-\dot{E}(r) = -K\frac{1}{(4\pi\epsilon_0)^2}\frac{4e^6}{\epsilon_0 c^3 m^2 r^4} \qquad (3.6)

Here negative sign indicates that the energy of the electron is decreasing. Combining Eqs. (3.4) and (3.6)

r^2\,dr = -\frac{K4e^4}{(4\pi\epsilon_0)\epsilon_0 c^3 m^2}dt \qquad (3.7)

Integrating the equation

\int_{10^{-10}}^{10^{-14}} r^2\,dr = -\frac{K4e^4}{(4\pi\epsilon_0)\epsilon_0 c^3 m^2}\int_0^T dt \qquad (3.8)

which yields

T \sim \frac{10^{-30}}{48\pi}\frac{(4\pi\epsilon_0)^2\,c^3 m^2}{Ke^4} \sim 5.26 \times 10^{-11}\,\mathrm{s}. \qquad (3.9)

Оригинал в Архива: IOQP2021-PartII-Questions-en.pdf · официални решения: IOQP2021-PartII-Solutions-20210217.pdf