Question Video: Identifying a Decay Process That Increases Atomic Number | Nagwa Question Video: Identifying a Decay Process That Increases Atomic Number | Nagwa

# Question Video: Identifying a Decay Process That Increases Atomic Number

Which of the following decay processes increases the atomic number of an atom? [A] ๐ฝโป decay [B] Positron emission [C] Electron capture [D] ๐ผ decay [E] ๐พ emission.

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### Video Transcript

Which of the following decay processes increases the atomic number of an atom? A) ๐ฝ minus decay, B) Positron emission, C) Electron capture, D) ๐ผ decay, or E) ๐พ 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, weโre looking for a decay process that would increase the number of protons in the nucleus. Before we filter the answers, letโs review what each one means.

In ๐ฝ decay, a neutron in a nucleus decays to form a proton, a high energy electron or ๐ฝ 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 ๐ฝ particles written this way, with an effective mass number of zero and effective atomic number of minus one. It isnโt 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 ๐ฝ sign helps to indicate that itโs a high-energy electron as a result of a nuclear decay.

Letโs move on to positron emission. Positron emission is otherwise known as ๐ฝ plus decay because another name for the positron is the ๐ฝ plus particle. Positron emission involves the complementary process to ๐ฝ minus decay. A proton decays to form a neutron, a positron, and an electron neutrino. The positron can be written as ๐ฝ 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 ๐ฝ plus particle. Instead, it reflects that a ๐ฝ 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 ๐ผ decay, a radioactive nucleus decays to form a smaller nucleus plus an ๐ผ particle. An ๐ผ 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 ๐ผ particle. Itโs fairly common to emit charges from such processes, so you may see the same process where the charge is not balanced.

Finally, ๐พ emission. ๐พ emission involves an excited nucleus, usually the product of a previous decay process, decaying into a stable nucleus plus a ๐พ 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 ๐พ rays. Now that weโve 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 ๐ฝ 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, ๐ฝ decay would increase the atomic number by one. Therefore, itโs our correct answer.

But, letโs 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 ๐ผ decay, the nucleus has a reduction in its atomic number of two. Two protons leave in the departing ๐ผ particle and there are no replacements. And finally, ๐พ emission. We see no reduction and no increase in the number of protons in the nucleus. Energy is lost as a ๐พ 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 ๐ฝ minus decay.

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