The discovery of the electron. Thomson measured the charge-to-mass ratio of the particles within cathode rays and it was -1.76 x 10^8 coulombs per gram. The mass of an electron 9.10 x 10^(-28) g. The dense nucleus contains over 99.9% of the mass of the atom, but it occupies very little of its volume.

1. The Nuclear Atom
Dr. K. Shahzad Baig
Memorial University of Newfoundland
(MUN)
Canada
Petrucci, et al. 2011. General Chemistry: Principles and Modern Applications. Pearson Canada Inc., Toronto, Ontario.
Tro, N.J. 2010. Principles of Chemistry. : A molecular approach. Pearson Education, Inc.

2. The Discovery of the Electron
If an evacuated glass tube is equipped with two electrodes and a voltage is applied, the
glass opposite the negative electrode is observed to glow from electrons emitted from the
cathode. Electrons were first discovered as the constituents of cathode rays.

3. Cathode rays are emitted by the negatively charged electrode, called the cathode, and
travel to the positively charged electrode, called the anode
Thomson found that these rays were actually streams of particles with the following
properties:
1. they traveled in straight lines;
2. they were independent of the composition of the material from which they originated
(the cathode); and
3. they carried a negative electrical charge.
Thomson measured the charge-to-mass ratio of the particles within cathode rays and it was
-1.76 x 108 coulombs per gram
Thomson had discovered the electron,
a negatively charged,
low mass particle present within all atoms.

4. Thomson’s mass-to-charge ratio for electrons, we can deduce the mass of an electron
as follows:
𝐶ℎ𝑎𝑟𝑔𝑒 𝑥
𝑚𝑎𝑠𝑠
𝑐ℎ𝑎𝑟𝑔𝑒
= 𝑚𝑎𝑠𝑠
−1.60 𝑥 10−19 𝐶 𝑥
𝑔
−1.76 𝑥 108 𝐶
= 𝑚𝑎𝑠𝑠
= 9.10 x 10-28 g
this mass is about 2000 times lighter than hydrogen

5. The principal types of radiation emitted by
radioactive substances are identified
alpha (α),
beta (β) particles,
gama (γ) rays [accidental discoveries]
X-Rays and Radioactivity

6. The Nuclear Atom
Thomson Model Rutherford Model
the nuclear atom a tiny, but massive, positively charged nucleus surrounded by lightweight, negatively
charged electrons

7. The Structure of the Atom
Rutherford proposed the nuclear theory of the atom, with three basic parts:
1. Most of the atom’s mass and all of its positive charge are contained in a small core
called the nucleus.
2. Most of the volume of the atom is empty space, throughout which tiny, negatively
charged electrons are dispersed.
3. There are as many negatively charged electrons outside the nucleus as there are
positively charged particles (named protons) within the nucleus, so that the atom is
electrically neutral.
The dense nucleus contains over 99.9% of the mass of the atom, but it occupies very little
of its volume.

8. Atomic Mass
𝐴𝑡𝑜𝑚𝑖𝑐 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑛 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
=
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙
𝑎𝑏𝑢𝑛𝑑𝑎𝑚𝑐𝑒 𝑜𝑓
𝑖𝑠𝑜𝑡𝑜𝑝𝑒 1
𝑥
𝑀𝑎𝑠𝑠 𝑜𝑓
𝑖𝑠𝑜𝑡𝑜𝑝𝑒 1
+
𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛𝑎𝑙
𝑎𝑏𝑢𝑛𝑑𝑎𝑚𝑐𝑒 𝑜𝑓
𝑖𝑠𝑜𝑡𝑜𝑝𝑒 2
𝑥
𝑀𝑎𝑠𝑠 𝑜𝑓
𝑖𝑠𝑜𝑡𝑜𝑝𝑒 2
+ …
The atomic mass (weight) of an element is a
weighted average based on an assigned value
of exactly 12 u for the isotope carbon-12.
This weighted average is measured
experimentally.
The precise mass of its atoms can be established
with a mass spectrometer.

9. Properties of Protons, Neutrons, and Electrons
at. mass of naturally occurring carbon = 0.9893 * 12 u + (1 - 0.9893) * 13.0033548378 u
= 13.0033548378 u - 0.9893 x (13.0033548378 u – 12 u)
= 13.0033548378 u - 0.9893 x (1.0033548378 u)
= 12.01 u

10. The Weighted-Average Atomic Mass
Example
The two naturally occurring isotopes of lithium, lithium-6 and lithium-7, have masses of
6.01512 u and 7.01600 u, respectively. Which of these two occurs in greater abundance?
Solve
From a table of atomic masses, the atomic mass of lithium is 6.941 u. Because this
value a weighted-average atomic mass is much closer to 7.01600 u than to 6.01512 u,
lithium-7 must be the more abundant isotope.

11. For instance, Li has Li7 and Li6 . If abundance of Li7 is (x), then Li6 abundance is
(1-x). So, the average atomic mass of lithium is:
The average atomic mass of whatever element calculates as sum of isotopes mass
multiplied by percentage of abundance on the Earth.
7x + 6(1-x)=6.941.
Let’s calculate abundance of both isotopes:
7x + 6 - 6x = 6.941
1x = 0.941
Li6 = 5.9%
Li7 = 94.1%
Li7 is most abundant.

12. PRACTICE EXAMPLE A:
The two naturally occurring isotopes of boron, boron-10 and boron-11, have masses of
10.0129370 u and 11.0093054 u, respectively. Which of these two occurs in greater
abundance?
PRACTICE EXAMPLE B:
Indium has two naturally occurring isotopes and a weighted atomic mass of 114.818 u.
One of the isotopes has a mass of 112.904058 u. Which of the following must be the
second isotope: or Which of the two naturally occurring isotopes must be the more
abundant?

13. EXAMPLE
Relating the Masses and Natural Abundances of Isotopes to the Atomic Mass of an
Element
Bromine has two naturally occurring isotopes. One of them, bromine-79, has a mass of
78.9183 u and an abundance of 50.69%. What must be the mass and percent natural
abundance of the other, bromine-81?
Solve
The atomic mass of Br is a weighted average of the masses of 79Br and 81Br:

14. Because the percent natural abundances must total 100%, the percent natural abundance
of 81Br is 100% - 50.69% = 49.31%.
79.904 u for the atomic mass [from periodic table ]
, 78.9183 u for the mass of 79Br, and the fractional abundances of the two isotopes, we
obtain
To four significant figures, the natural abundance of the bromine-81 isotope is 49.31%
and its mass is 80.92 u.

15. Problem Statement
With mass spectral data, the mass of an oxygen-16 atom is found to be 1.06632 times that
of a nitrogen-15 atom. Given that 16O has a mass of 15.9949 u (see above), what is the
mass of a nitrogen-15 atom, in u?
Solution
𝑚𝑎𝑠𝑠 𝑜𝑓 16𝑂
𝑚𝑎𝑠𝑠 𝑜𝑓 15𝑁
= 1.06632
We solve the expression above for the mass of 15N and then substitute 15.9949 u for the
mass of 16O. We obtain the result
= mass of 15N =
𝑚𝑎𝑠𝑠 𝑜𝑓 16𝑂
1.06632
=
15.9949
1.06632
= 15.0001 𝑢

16. PRACTICE EXAMPLE A:
The masses and percent natural abundances of the three naturally occurring isotopes of
silicon are 27.9769265325 u, 92.223%; 28.976494700 u, 4.685%; 29.973377017 u,
3.092%. Calculate the weighted-average atomic mass of silicon
PRACTICE EXAMPLE A:
The two naturally occurring isotopes of boron, boron-10 and boron-11, have masses of
10.0129370 u and 11.0093054 u, respectively. Which of these two occurs in greater
abundance?
PRACTICE EXAMPLE B:
Indium has two naturally occurring isotopes and a weighted atomic mass of 114.818 u.
One of the isotopes has a mass of 112.904058 u. Which of the following must be the
second isotope: or Which of the two naturally occurring isotopes must be the more
abundant?

17. Properties of Protons, Neutrons, and Electrons
An individual atom is characterized in terms of its atomic number (proton number) Z
and mass number, A.
The difference, (A-Z) , is the neutron number.
The masses of individual atoms and their component parts are expressed in atomic
mass units (u).
All elements from Z =1 to Z = 112 have been characterized and all but element 112
have been given a name and chemical symbol. Knowledge of the several elements
following Z = 112 is more tenuous.
Isotopes are atoms of the same element that differ in mass number