# Nuclear Reactions

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## Homework Statement

Explain why deuterium cannot undergo ##\beta## decay or produce a stable nucleus while tritium can.

## Homework Equations

Deuterium : ##{^2_1}{H}##
Tritium : ##{^3_1}{H}##

## The Attempt at a Solution

Observe the negative ##\beta## decay of deuterium :
$${^2_1}{H} \stackrel{\beta^-}{\rightarrow} {^2_2}{He} + {^0_{-1}}{e}$$

The positive ##\beta## decay of deuterium :
$${^2_1}{H} \stackrel{\beta^+}{\rightarrow} {^2_0}{He} + {^0_{1}}{e}$$

The negative ##\beta## decay of tritium :
$${^3_1}{H} \stackrel{\beta^-}{\rightarrow} {^3_2}{Li} + {^0_{-1}}{e}$$

The positive ##\beta## decay of tritium :
$${^3_1}{H} \stackrel{\beta^+}{\rightarrow} {^3_0}{Li} + {^0_{1}}{e}$$

When deuterium undergoes ##\beta## decay, it either produces an unstable nucleus or a nucleus with no protons. With no protons, there would be nothing holding the atom together so the whole thing would fall apart.

Tritium seems to be able to undergo a stable ##\beta^-## decay and produce a stable nucleus. The positive decay of tritium destroys the nucleus entirely.

Is this reasoning okay? The question is a bit vague.

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SteamKing
Staff Emeritus
Homework Helper
You're confused on your atomic reactions. Lithium does not have 2 protons, it has 3. Helium has 2 protons. If a reaction results in the loss of all protons from the nucleus, then show only the resulting atomic particles on the RHS. Don't make up nuclei that don't exist.

Homework Helper
You're confused on your atomic reactions. Lithium does not have 2 protons, it has 3. Helium has 2 protons. If a reaction results in the loss of all protons from the nucleus, then show only the resulting atomic particles on the RHS. Don't make up nuclei that don't exist.

I mixed my numbers up a bit there. So both of the negative beta decays will cause helium to be formed.

The positive beta decays destroy the nuclei in both cases, so all that would be left in both cases would be the electron.

SteamKing
Staff Emeritus
Homework Helper
I mixed my numbers up a bit there. So both of the negative beta decays will cause helium to be formed.

The positive beta decays destroy the nuclei in both cases, so all that would be left in both cases would be the electron.

In the positron decays, the nuclei aren't annihilated, but they do leave some atomic remnants behind in addition to the electrons. Your reactions must show a balance in the total atomic mass units on both sides of the equation.

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So these two are okay :

The negative ##\beta## decay of deuterium :
$${^2_1}{H} \stackrel{\beta^-}{\rightarrow} {^2_2}{He} + {^0_{-1}}{e}$$

The negative ##\beta## decay of tritium :
$${^3_1}{H} \stackrel{\beta^-}{\rightarrow} {^3_2}{Li} + {^0_{-1}}{e}$$

These two should be like so now :

The positive ##\beta## decay of deuterium :
$${^2_1}{H} \stackrel{\beta^+}{\rightarrow} {^2_0}{n} + {^0_{1}}{e}$$

The positive ##\beta## decay of tritium :
$${^3_1}{H} \stackrel{\beta^+}{\rightarrow} {^3_0}{n} + {^0_{1}}{e}$$

Where I use ##n## to denote a neutron. I'm not quite seeing why deuterium is unstable compared to tritium?

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SteamKing
Staff Emeritus
Homework Helper
You're still having problems with the atomic notation. The lower number reflects the number of protons (also called the atomic number Z) in the particle or nucleus. The upper number reflects the total number of atomic mass units (amu) in the particle or nucleus.

A neutron has zero protons and has a mass of 1 amu (more or less); therefore it is denoted by

01n (pardon the offset scripts)

If you have more than one neutron, then you show this by adding a normal prefix (just like for atoms in a chemical equation). Thus, 3 neutrons are shown:

3 01n

Often times, for clarity and ease of writing, common atomic particles are written thus:

p - proton
n - neutron
e - electron
e+ - positron
α - alpha particle (helium-4 nucleus)
β - beta particle (electron emitted from the nucleus)
$\gamma$ - gamma ray

This worksheet explains basic atomic notation:
http://web.gccaz.edu/~ldiebolt/130ll/nuclearchem.pdf

Again, you are showing lithium having an atomic number of 2 when it has an atomic number of 3.
Your reaction showing tritium decaying into lithium is incorrect. If anything, tritium is going to decay into Helium-3.

• 1 person
Homework Helper
You're still having problems with the atomic notation. The lower number reflects the number of protons (also called the atomic number Z) in the particle or nucleus. The upper number reflects the total number of atomic mass units (amu) in the particle or nucleus.

A neutron has zero protons and has a mass of 1 amu (more or less); therefore it is denoted by

01n (pardon the offset scripts)

If you have more than one neutron, then you show this by adding a normal prefix (just like for atoms in a chemical equation). Thus, 3 neutrons are shown:

3 01n

Often times, for clarity and ease of writing, common atomic particles are written thus:

p - proton
n - neutron
e - electron
e+ - positron
α - alpha particle (helium-4 nucleus)
β - beta particle (electron emitted from the nucleus)
$\gamma$ - gamma ray

This worksheet explains basic atomic notation:
http://web.gccaz.edu/~ldiebolt/130ll/nuclearchem.pdf

Again, you are showing lithium having an atomic number of 2 when it has an atomic number of 3.
Your reaction showing tritium decaying into lithium is incorrect. If anything, tritium is going to decay into Helium-3.

That article was a bit more clear than what I was reading for sure, thanks for finding it.

I copied my first post and I accidentally didn't change the Li to He. Totally missed that. Here's the fixed reactions :

The ##\beta^-## decays :

The negative ##\beta## decay of deuterium :
$${^2_1}{H} \stackrel{\beta^-}{\rightarrow} {^2_2}{He} + {^0_{-1}}{e}$$

The negative ##\beta## decay of tritium :
$${^3_1}{H} \stackrel{\beta^-}{\rightarrow} {^3_2}{He} + {^0_{-1}}{e}$$

I think I see it now. The deuterium decays and will have zero neutrons, one negatively charged electron and two protons. Without any neutrons the reaction is completely unstable.

The tritium at least has a neutron floating around to keep things more stable than in the deuterium reaction.

The ##\beta^+## decays :

The positive ##\beta## decay of deuterium :
$${^2_1}{H} \stackrel{\beta^+}{\rightarrow} 2{^1_0}{n} + {^0_{1}}{e}$$

The positive ##\beta## decay of tritium :
$${^3_1}{H} \stackrel{\beta^+}{\rightarrow} 3{^1_0}{n} + {^0_{1}}{e}$$

Neither of these appear stable at all? I guess I should only be concerned with the ##\beta^-## decays.

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