Video Transcript
Which of the following decay
processes increases the atomic number of an atom? A) ๐ฝ minus decay, B) Positron
emission, C) Electron capture, D) ๐ผ decay, or E) ๐พ emission.
All five processes refer to changes
in the nucleus of an atom or ion. So, you can call these nuclear
processes. The atomic number of an atom is the
number of protons in the nucleus of that atom. So, weโre looking for a decay
process that would increase the number of protons in the nucleus. Before we filter the answers, letโs
review what each one means.
In ๐ฝ decay, a neutron in a nucleus
decays to form a proton, a high energy electron or ๐ฝ particle, and an
antineutrino. This is how we write it in symbol
notation. A single neutron has a mass number
of one and atomic number of zero. A single proton has a mass number
of one as well, but an atomic number of one. You may see ๐ฝ particles written
this way, with an effective mass number of zero and effective atomic number of minus
one. It isnโt actually possible to have
a negative atomic number. The minus one is introduced to make
it easier to balance the equation. The sum of the atomic numbers on
the left is equal to the sum of the effective atomic numbers on the right. You may see the electrons simply
written as e minus. The ๐ฝ sign helps to indicate that
itโs a high-energy electron as a result of a nuclear decay.
Letโs move on to positron
emission. Positron emission is otherwise
known as ๐ฝ plus decay because another name for the positron is the ๐ฝ plus
particle. Positron emission involves the
complementary process to ๐ฝ minus decay. A proton decays to form a neutron,
a positron, and an electron neutrino. The positron can be written as ๐ฝ
plus, with an effective atomic number of one to help with balancing the
equation. Again, this number does not reflect
the number of protons in the ๐ฝ plus particle. Instead, it reflects that a ๐ฝ plus
particle is the result of a decay of a proton. You may see positrons written
simply as e plus.
Electron capture is a rare process
where a nucleus absorbs an electron from the inner shell, turning a proton into a
neutron and an electron neutrino. This is how we might write it in
symbols. In ๐ผ decay, a radioactive nucleus
decays to form a smaller nucleus plus an ๐ผ particle. An ๐ผ particle is the equivalent of
a helium nucleus. We can show the process
symbolically by taking a nucleus X, which loses four from its mass number, two from
its atomic number, forming a smaller nucleus and an ๐ผ particle. Itโs fairly common to emit charges
from such processes, so you may see the same process where the charge is not
balanced.
Finally, ๐พ emission. ๐พ emission involves an excited
nucleus, usually the product of a previous decay process, decaying into a stable
nucleus plus a ๐พ ray. One example of this will be the
decay of an excited thorium-234 atom. Upon decay to its ground state, the
excited thorium releases two ๐พ rays. Now that weโve reviewed all five
decay processes, we need to find which one would increase, rather than keep the same
or decrease, the atomic number of the nucleus.
Firstly, in ๐ฝ decay, we see a
neutron transform into a proton. This would increase the atomic
number of the nucleus because both particles will be in the nucleus. So, ๐ฝ decay would increase the
atomic number by one. Therefore, itโs our correct
answer.
But, letโs go through the others,
just in case. In positron emission, a proton is
consumed, leaving a neutron behind. So, the atomic number will actually
decrease rather than increase. Similarly, in electron capture, a
proton is converted into a neutron. So, we again have an atomic number
decrease of one. For an ๐ผ decay, the nucleus has a
reduction in its atomic number of two. Two protons leave in the departing
๐ผ particle and there are no replacements. And finally, ๐พ emission. We see no reduction and no increase
in the number of protons in the nucleus. Energy is lost as a ๐พ ray, but
there is no change to the number of protons in the nucleus. So, of the five decay processes
given, the only one that increases the atomic number of the associated atom is ๐ฝ
minus decay.