How do you write a balanced nuclear equation for alpha decay of Po-218?
Here's how you can do that.
When a radioactive nuclide undergoes alpha decay, it emits an alpha particle,
This means that after the alpha particle is emitted
- the mass number of the nuclide will decrease by
#4# #-># this happens because the alpha particle contains#2# protons and#2# neutrons- the atomic number of the nuclide will decrease by
#2# #-># this happens because the alpha particle contains#2# protons
You can thus say that you have
#""_ (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 -># conservation of charge
is lead,
#218 - 4 = 214 -># conservation of mass
The balanced nuclear equation that describes the alpha decay of polonium-218 will look like this
#""_ (color(white)(.)84)^218"Po" -> ""_ (color(white)(.)82)^214"X" + ""_ 2^4alpha#
As
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[_{84}^{218}\text{Po} \rightarrow ,_2^4\text{He} + ,_82^{214}\text{Pb}]
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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