Lithium Decay

Chemistry Level 1

3 5 L i → 2 4 H e + ? ‾ \large _3 ^5 \ce{Li -> } _2^4 \ce{He } \, + \underline ?

Lithium-5 undergoes a form of radioactive decay to form Helium-4 through the release of a particle. What is this particle?

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Proton Neutron Electron Positron

David Hontz by David Hontz , 26, USA

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3 solutions

David Hontz
Jun 30, 2016

3 5 L i → 2 4 H e + ? ‾ 3 − 2 5 − 4 ? = 1 1 p Answer ⇒ p r o t o n _3^5Li \rightarrow _2^4He + \underline{?} \\ _{3-2}^{5-4}? = _1^1p \\ \text{Answer} \Rightarrow \boxed{proton}

8 Helpful 0 Interesting 0 Brilliant 0 Confused
Leonblum Iznotded
Jul 29, 2018

(This is no chemistry, this is nuclear physics.) In chemical reaction, every species is preserved...

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Lithium-5 :

3 p + 3 p^{+} [ Z = 3 Z=3 ]

2 n 0 2 n^{0} [Z-A=5-2]

(The number of electrons is insignificant.)

It decays to form Helium-4 after a particle has been released.

Breaking down Helium-4, we get:

2 p + 2 p^{+} (atomic number 2)

2 n 0 2 n^{0} (atomic mass 4 - atomic number 2)

∴ 1 p r o t o n \therefore \boxed{1 proton} has been lost in this case.

1 Helpful 1 Interesting 0 Brilliant 0 Confused

Actually, the number of electrons is NOT insignificant, an electron balances the charge of the proton, so when a proton is ejected, and electron has to go with it as well.

Robert Cormia - 2 years, 3 months ago

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