How do you write a balanced nuclear equation for alpha decay of Po-218?

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How do you write a balanced nuclear equation for alpha decay of Po-218?

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Answer 1 · 2017-07-11T23:47:20

Here's how you can do that.

Explanation:

When a radioactive nuclide undergoes alpha decay, it emits an alpha particle, $_{2}^{4} α$ $""_2^4alpha$ , which is essentially the nucleus of a helium-4 atom.

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This means that after the alpha particle is emitted

the mass number of the nuclide will decrease by $4$ $4$ $\to$ $->$ this happens because the alpha particle contains $2$ $2$ protons and $2$ $2$ neutrons

the atomic number of the nuclide will decrease by $2$ $2$ $\to$ $->$ this happens because the alpha particle contains $2$ $2$ protons

You can thus say that you have

$_{. 84}^{218} Po \to_{. (84 - 2)}^{(218 - 4)} X +_{2}^{4} α$ $""_ (color(white)(.)84)^218"Po" -> ""_ (color(white)(.)(84-2))^((218 - 4))"X" + ""_ 2^4alpha$

A quick look in the Periodic Table of Elements will show that the element that has the atomic number equal to

$84 - 2 = 82 \to$ $84 - 2 = 82 ->$ conservation of charge

is lead, $Pb$ $"Pb"$ . This means that the resulting nuclide will be lead-214 since its mass number is equal to

$218 - 4 = 214 \to$ $218 - 4 = 214 ->$ conservation of mass

The balanced nuclear equation that describes the alpha decay of polonium-218 will look like this

$_{. 84}^{218} Po \to_{. 82}^{214} X +_{2}^{4} α$ $""_ (color(white)(.)84)^218"Po" -> ""_ (color(white)(.)82)^214"X" + ""_ 2^4alpha$

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How do you write a balanced nuclear equation for alpha decay of Po-218?

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