Faculty of Engineering Sciences
Quiz 1
Student name
ELE 163
Department of Basic sciences
Electrical material
17-3-2013
Time: 45 min
___________________________________________________________________
Q1 Circle the correct answer
( 5 marks )
1- Covalently bonding normally occurs between which types of elements?
A metals only.
B non-metals only.
C metals and non-metals.
2. Which one of the following is not a strong bond?
A van der Waals bond
B Covalent bond
C Metallic bond
D none of these.
D Ionic bond
3- Cat ions are :
A Negatively charged
B positively charged
C neutral
D none of the above
4- The energy released when an atom takes an electron is called:
A
internal energy
D electron affinity
B.
E.
dissociation energy
ionization energy
C
binding energy
5- Metallic bond is not characterized by :
A ductility.
B high conductivity.
C Non directionality.
Q2 Solve the following question :
A- 1.Write and prove Virial theorem?
5 marks
‫اجب على ظهر الورقة‬
1
D Bad conductivity
2.Write the electronic structure of the following elements:
20Ca
:1s2 2s2 2p6 3s2 3p6 4s2
17Cl
: 1s2 2s2 2p6 3s2 3p5
11Na
: 1s2 2s2 2p6 3s1
9F
20Ca, 11Na, 17Cl,
and 9F
: 1s2 2s2 2p5
Q3 Solve the following problem:
5 marks ‫اجب على ظهر الورقة‬
Consider the H2 molecule in a simple way as two touching H atoms as depicted in the Figure. The radius ro of the
hydrogen atom is 0.0529 nm, the PE in the isolated H-atom is (-27.2 eV), me = 9.1x10-31kg, e=1.6x10-19 C ,k= 9x109
Nm2/C2
A. Calculate the total energy of this arrangement Does this arrangement have a lower energy than two separated H
atoms?
B. Calculate the change in PE in going from two isolated H-atoms to the H2 molecule. Using the Virial theorem, find the
change in the total energy and hence the covalent bond energy?
A. Consider the PE of the whole arrangement of charges shown in the figure. In evaluating the PE of all the charges, we must avoid
double counting of interactions between the same pair of charges. The total PE is the sum of the following:
Electron 1 interacting with the proton at a distance ro on the left, with the proton at ro on the right and with electron 2 at a
distance 2ro
+ Electron 2 on the far left interacting with a proton at ro and another proton at 3ro
+ Two protons, separated by 2ro, interacting with each other
PE  
e2
e2
e2


4 o ro 4 oro 4 o (2ro )

2
2
e
e

4 o ro 4 o 3ro

e2
4  o 2ro
Substituting and calculating, we find PE = -1.0176  10-17 J or -63.52 eV
The negative PE for this particular arrangement indicates that this arrangement of charges is indeed energetically favorable compared
with all the charges infinitely separated (PE is then zero).
B. The potential energy of an isolated H-atom is -2 13.6 eV or -27.2 eV. The difference between the PE of the H2 molecule and two
isolated H-atoms is,
PE = - (63.52) eV - 2(-27.2) eV=9.12eV
We can write the last expression above as the change in the total energy.
E 
1
1
 PE  (9.12eV)  4.56eV
2
2
This change in the total energy is negative. The H2 molecule has lower energy than two H-atoms by 4.56 eV which is the
bonding energy. This is very close to the experimental value of 4.51 eV
2
a. Consider the PE of the whole arrangement of charges shown in the figure. In evaluating the PE of all the
charges, we must avoid double counting of interactions between the same pair of charges. The total PE is the
sum of the following:
Electron 1 interacting with the proton at a distance ro on the left, with the proton at ro on the right and
with electron 2 at a distance 2ro
+ Electron 2 on the far left interacting with a proton at ro and another proton at 3ro
+ Two protons, separated by 2ro, interacting with each other
3
PE  
e2
e2
e2


4 o ro 4 oro 4 o (2ro )
2
2
e
e


4 o ro 4 o 3ro

e2
4  o 2ro
Substituting and calculating, we find PE = -1.0176  10-17 J or -63.52 eV
The negative PE for this particular arrangement indicates that this arrangement of charges is indeed
energetically favorable compared with all the charges infinitely separated (PE is then zero).
b. The potential energy of an isolated H-atom is -2 13.6 eV or -27.2 eV. The difference between the PE of the
H2 molecule and two isolated H-atoms is,
PE = - (63.52) eV - 2(-27.2) eV=9.12eV
We can write the last expression above as the change in the total energy.
E 
1
1
 PE  (9.12eV)  4.56eV
2
2
This change in the total energy is negative. The H2 molecule has lower energy than two H-atoms by 4.56
eV which is the bonding energy. This is very close to the experimental value of 4.51 eV
4