How do you write an equation for the beta decay of sodium-24?

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How do you write an equation for the beta decay of sodium-24?

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Answer 1 · 2017-06-16T14:05:59

Here's how you can do that.

Explanation:

Sodium-24 undergoes beta decay, or, more specifically, beta-minus decay.

During a beta decay, a neutron located in the nucleus of a radioactive nuclide is converted to a proton. At the same time, the nuclide emits an electron, also called a beta particle, and an electron antineutrino, ${¯ ν}_{e}$ $bar(nu)_"e"$ .

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Since a neutron is being converted to a proton, you can say that the atomic number of the nuclide increases by $1$ $1$ and the mass number of the nuclide remains unchanged.

You will thus have

$_{11}^{24} Na \to_{Z}^{A} ? +_{- 1}^{- 0} β + {¯ ν}_{e}$ $""_ 11^24"Na" -> ""_ Z^A"?" + ""_ (-1)^(color(white)(-)0)beta + bar(nu)_"e"$

In any nuclear reaction, charge and mass must be conserved, so you have

$24 = A + 0 \to$ $24 = A + 0 ->$ conservation of mass

This will get you

$A = 24 \to$ $A = 24 ->$ the mass number remains unchanged!

$11 = Z + (- 1) \to$ $11 = Z + (-1) ->$ conservation of charge

This will get you

$Z = 12 \to$ $Z = 12 ->$ he atomic number increases by $1$ $1$ !

Therefore, you can say that the resulting nuclide will have $Z = 12$ $Z = 12$ and $A = 24$ $A = 24$ . A quick look in the Periodic Table will reveal that you're dealing with magnesium-24.

This means that you have

$_{11}^{24} Na \to_{12}^{24} Mg +_{- 1}^{- 0} β + {¯ ν}_{e}$ $""_ 11^24"Na" -> ""_ 12^24"Mg" + ""_ (-1)^(color(white)(-)0)beta + bar(nu)_"e"$

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How do you write an equation for the beta decay of sodium-24?

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