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INPhO 2020, Теория — Задача 3. Bohr model and ionisation

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Всички материали от INPhO 2020

Indian National Physics Olympiad – 2020 · 2 февруари 2020 г. · 15 т.

Внимание: Бележка към темата

The complete retained question booklet includes the summary answer boxes, sixteen blank detailed-answer pages and two blank graph sheets. All matching archived HBCSE key content is retained, including its handwritten graph and comment not for grading purposes. Printed sign/symbol discrepancies and the key cover page-count mismatch remain literal with labelled source notes.

Native cover and instructions

Indian National Physics Olympiad – 2020

Date: 02 February 2020. Time: 09:00–12:00 (3 hours). Maximum Marks: 80.

Roll Number: $2\,0\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ . Extra sheets attached: [blank box]. INO Centre (e.g. Ranchi): [blank line].

(Do not write below this line)

This booklet consists of 23 pages (excluding this page) and total of 5 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. 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.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets (maximum two) from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Constant	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	13
2	12
3	15
4	20
5	20
Total	80

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key notice

Please note that alternate/equivalent methods and different way of expressing final solutions may exist.

Source note: The retained key cover says “16 pages (excluding this page)”; the native PDF contains eighteen pages, including its appended handwritten graph. Its cover otherwise repeats the question cover instructions and constants.

Условие

Consider the Bohr model of the hydrogen atom. Let $m_e$ and $e$ be the mass and magnitude of the charge of the electron respectively. Let $a_0$ be the ground state radius (Bohr radius).

a) Obtain an expression for the ionisation energy $I_{\mathrm H}$ of the ground state of the hydrogen atom in terms of $a_0$ and constants.

Summary answer box: $I_{\mathrm H}=$ [blank]. [1 т.]

b) Consider a singly ionised helium atom $\mathrm{He}^+$ . Obtain the ground state ionisation energy $I_{\mathrm{He}^+}$ of $\mathrm{He}^+$ in terms of $I_{\mathrm H}$ .

Summary answer box: $I_{\mathrm{He}^+}=$ [blank]. [2 т.]

c) Now consider a two electron system with arbitrary atomic number $Z$ . Use the Bohr model to obtain the ground state radius ( $r(Z)$ ) in terms of $a_0$ and $Z$ . Assume the two electrons are in the same circular orbit and as far apart as possible.

Summary answer box: $r(Z)=$ [blank]. [1 т.]

d) Derive an expression for the first ionisation energy $I_Z^{\mathrm{th}}$ for two electron system with arbitrary $Z$ in terms of $Z$ and $I_{\mathrm H}$ .

Summary answer box: $I_Z=$ [blank]. [3 т.]

e) The table below contains the experimental data of $I_Z^{\mathrm{expt}}$ (in units of Rydberg where $1\ \mathrm{Ryd}=13.6\ \mathrm{eV}$ ) versus $Z$ for various two-electron systems.

	$Z$	$I_Z^{\mathrm{expt}}$
$\mathrm H^-$	1	0.055
He	2	1.81
$\mathrm{Li}^+$	3	5.56
$\mathrm{Be}^{++}$	4	11.32
$\mathrm B^{3+}$	5	19.07
$\mathrm C^{4+}$	6	28.83
$\mathrm N^{5+}$	7	40.60
$\mathrm O^{6+}$	8	54.37
$\mathrm F^{7+}$	9	70.15

Experimental values were not found to be equal to the theoretical predictions. This difference arises mainly from non-inclusion of Pauli’s principle in the theoretical derivation of part (d). It was suggested that if the value of $Z$ was reduced by some fixed amount $\alpha$ ( $Z^*=Z-\alpha$ ) in the final expression of $I_Z^{\mathrm{th}}$ obtained in part (d), then $I_{Z^*}^{\mathrm{th}}\approx I_Z^{\mathrm{expt}}$ . Draw a suitable linear plot and from the graph find $\alpha$ . Two graph papers are provided with this booklet in case you make a mistake.

Whole native blank green graph-paper printedpage21, header and question-number box and entire original grid.

Whole second native blank green graph-paper printedpage22, header/question-number/roll field and entire original grid.

Summary answer box: $\alpha=$ [blank].

Detailed answers can be found on page numbers: [blank].

Source note: The native printed page3 repeats the “Detailed answers can be found on page numbers:” line twice below this task. Both are blank response fields. [8 т.]

Решение

(a)

Centripetal acceleration is given by Coulomb force.

\frac{m_ev^2}{a_0}=\frac{Ke^2}{a_0^2};\quad K=\frac1{4\pi\epsilon_0};\quad m_evr=\hbar\qquad (3.1)

a_0=\frac{\hbar^2}{Km_ee^2}\qquad (3.2)

\text{Total energy}=\frac{m_ev^2}{2}-\frac{Ke^2}{a_0}=-\frac{Ke^2}{2a_0}\qquad (3.3)

I_{\mathrm H}=\frac{Ke^2}{2a_0}\qquad (3.4)

(b)

\frac{m_ev^2}{r}=\frac{2Ke^2}{r^2}\text{ and }m_evr=\hbar\qquad (3.5)

r=\frac{\hbar^2}{2Km_ee^2}=\frac{a_0}{2}\qquad (3.6)

\text{Total Energy (T.E.)}=\frac12m_ev^2-\frac{2Ke^2}{r}=-\frac{Ke^2}{r}=\frac{2Ke^2}{a_0}=-4I_{\mathrm H}\qquad (3.7)

I_{\mathrm{He}^+}=4I_{\mathrm H}\qquad (3.8)

(c)

\frac{m_ev^2}{r}=\frac{KZe^2}{r^2}-\frac{Ke^2}{(2r)^2}\text{ and }m_evr=\hbar\qquad (3.9)

m_ev^2r=Ke^2\left(Z-\frac14\right)\text{ and }m_ev^2r=\frac{\hbar^2}{m_er}\qquad (3.10)

r=\frac{\hbar^2}{Km_ee^2\left(Z-\frac14\right)}=\frac{a_0}{\left(Z-\frac14\right)}\qquad (3.11)

(d)

\text{Kinetic Energy (K.E.)}=m_ev^2=\frac{Ke^2\left(Z-\frac14\right)}{r}\qquad (3.12)

=\frac{Ke^2\left(Z-\frac14\right)^2}{a_0}\qquad (3.13)

=2\left(Z-\frac14\right)^2I_{\mathrm H}\qquad (3.14)

\text{Potential Energy (P.E.)}=-\frac{2KZe^2}{r}+\frac{Ke^2}{2r}\qquad (3.15)

=\frac{-2Ke^2}{r}\left(Z-\frac14\right)\qquad (3.16)

=\frac{-2Ke^2}{a_0}\left(Z-\frac14\right)^2\qquad (3.17)

=-4\left(Z-\frac14\right)^2I_{\mathrm H}\qquad (3.18)

\mathrm{(T.E.)}_i=-2\left(Z-\frac14\right)^2I_{\mathrm H}\qquad (3.19)

\mathrm{(T.E.)}_f=-Z^2I_{\mathrm H}\qquad (3.20)

I_Z^{\mathrm{th}}=\mathrm{(T.E.)}_f-\mathrm{(T.E.)}_i=2\left(Z-\frac14\right)^2I_{\mathrm H}-Z^2I_{\mathrm H}\qquad (3.21)

I_Z^{\mathrm{th}}=\left(Z^2-Z+\frac18\right)I_{\mathrm H}\qquad (3.22)

(e)

\Delta I_Z=I_Z^{\mathrm{th}}-I_Z^{\mathrm{expt}}=I_Z^{\mathrm{th}}-I_{Z^*}^{\mathrm{th}}=\left\{\left[Z^2-Z+\frac18\right]-\left[(Z-\alpha)^2-(Z-\alpha)+\frac18\right]\right\}I_{\mathrm H}\qquad (3.23)

\Delta I_Z=\left[-\alpha^2+2Z\alpha-\alpha\right]I_{\mathrm H}\qquad (3.24)

	$Z$	$(I_Z^{\mathrm{expt}})$	$I_Z^{\mathrm{th}}$	$(I_Z^{\mathrm{th}}-I_Z^{\mathrm{expt}})$
$\mathrm H^-$	1	0.055	0.125	0.07
He	2	1.81	2.13	0.32
$\mathrm{Li}^+$	3	5.56	6.13	0.57
$\mathrm{Be}^{++}$	4	11.32	12.13	0.81
$\mathrm B^{3+}$	5	19.07	20.13	1.06
$\mathrm C^{4+}$	6	28.83	30.13	1.30
$\mathrm N^{5+}$	7	40.60	42.13	1.53
$\mathrm O^{6+}$	8	54.37	56.13	1.76
$\mathrm F^{7+}$	9	70.15	72.13	1.98

A plot of $\Delta I_Z$ vs $Z$ is a linear graph.

Slope of the graph $=2\alpha=0.24$

$\alpha=0.12$

One can also linearize in the following way:

I_Z^{\mathrm{expt}}=I_{Z^*}^{\mathrm{th}}=\left(Z-\alpha-\frac12\right)^2-\frac18\qquad (3.25)

\sqrt{I_Z^{\mathrm{expt}}+\frac18}=Z-\alpha-\frac12\qquad (3.26)

Plot of $\sqrt{I_Z^{\mathrm{expt}}+\frac18}$ vs $Z$ is a linear graph. This method gives $\alpha=0.10$ . Both methods are acceptable.

Accepted range of $\alpha$ is : $0.10\leq\alpha\leq0.14$ .

Graph is on the last page of this booklet.

Whole native handwritten INPhO2020 detailed-answer graph, all nine circledpoints/green grid/negativeintercept/allscaledaxes/scaleand slope arithmetic/nativeheader. No cleanup/redraw.

Source note: Equation(3.7) prints positive $2Ke^2/a_0$ within a chain ending at $-4I_{\mathrm H}$ ; its signs remain literal. Equations(3.25)–(3.26) omit an explicit $I_{\mathrm H}$ factor while the preceding table uses Rydberg values. The native formulas, both accepted methods and handwritten graph are preserved.

Оригинал в Архива: INPHO2020-Questions-en.pdf · решение от архива: INPHO2020-Solutions-20200220.pdf