How to write proper nuclear reactions?

Question

a) Beta-decay of radon-222

b) Alpha-decay of uranium-238

Answer 1

Before we delve into the details of the problem, let's review some notations!

The notation _Z^A​X represents a nucleus where:

1. A is the mass number (total number of protons and neutrons)
2. Z is the atomic number (number of protons)
3. X is the chemical symbol of the element

a) Beta-decay of radon-222

Remember, Beta decay occurs when a neutron in the nucleus transforms into a proton, emitting an electron (beta particle) and an antineutrino.

For radon-222, we have:

1. Initial nucleus: ₈₆²²²​Rn
2. Beta particle: β− (an electron)
3. Resulting nucleus: ₈₇²²²Fr (Francium-222)

Therefore, the nuclear reaction for beta-decay of radon-222 is

₈₆²²²​Rn → ​β− + ₈₇²²²Fr + ¯v_{e (supposed to be a v_bar_e that represents antineutrino, but it is hard to type here)}

b) Alpha-decay of uranium-238

Alpha decay occurs when the nucleus emits an alpha particle (which is a helium-4 nucleus).

For uranium-238:

1. Initial nucleus: ₉₂²³⁸U
2. Alpha particle: ₂⁴He
3. Resulting nucleus: ₉₀²³⁴Th (Thorium-234)

The nuclear reaction for alpha-decay of uranium-238 is:

₉₂²³⁸U→ ₉₀²³⁴Th + ₂⁴He

Let me know if you need further clarification! Hope you have a great day :)

Cheers,

Devany

To answer your first three questions, I will try to break down the logic behind the problem - but in short yes, you subtract the corresponding number of proton/neutron/... emitted due to the particle

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a) Beta-decay of radon-222

Starting Nucleus: Radon-222 has 86 protons and 136 neutrons (222 - 86 = 136)

Beta Decay Process: A neutron in the nucleus transforms into a proton, emitting an electron (beta particle) and an antineutrino

Resulting Nucleus: The atomic number increases by 1 (from 86 to 87 because a neutron is transformed into a proton), and the mass number remains the same (because the total number of neutrons and the proton is the same)

Thus, the resulting nucleus is Francium-222, which has 87 protons and 135 neutrons (222 - 87 = 135).

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b) Alpha-decay of uranium-238

Starting Nucleus: Uranium-238 has 92 protons and 146 neutrons (238 - 92 = 146)

Alpha Decay Process: The nucleus emits an alpha particle, which is a helium-4 nucleus (2 protons and 2 neutrons)

Resulting Nucleus: The atomic number decreases by 2 (from 92 to 90 because 2 protons are emitted), and the mass number decreases by 4 (from 238 to 234 because 4 protons and neutrons in total are emitted

Thus, the resulting nucleus is Thorium-234, which has 90 protons and 144 neutrons (234 - 90 = 144).

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Q: Does the average atomic mass generally tend to vary across the periodic table of elements? Thorium-232 vs Thorium234?

A: The atomic mass of an element listed in the periodic table is the weighted average of all its isotopes' masses based on their natural abundance. However, there are also different isotopes of an element, which means they have the same number of protons but different numbers of neutrons.

In our case, Thorium-232 and Thorium-234 are both isotopes of Thorium, but with different neutron counts. Thorium-232 is the one that is the most abundant and the most stable in nature, but Thorium-234 also exists :)

As I mentioned previously, Thorium-234 has 144 neutrons, which you can get by subtracting 234 by 90 (90 is the proton number, and it always stays the same for the same element). Keep in mind that, in the case of nuclear reactions, the specific isotopes involved are what matter.

Small exercise for you: for the naturally abundant Thorium-232, how many protons does it have? How would you write it?

I hope that helps!