Question

How many alpha particles are emitted in the series of radioactive decay events from a U-238 nucleus to a Pb-206 nucleus?

Answer 1

`"8 alpha decays which release 8 alpha particles"`

Explanation:

`U^238 rarr Pb^206`

(This is not a direct decay but is a process of many alpha decays and beta decay).

An alpha particle has 2neutrons and 2protons(A helium nucleus)

Thus the combined mass is 4

You only need how many alpha decay it undergoes . This can be determined by an equation like this

`206 = 238 - (4x)`

Solve for x

`238-4x-238=206-238`(Subtract 238 from both sides )

`-4x= -32`

Thus`-x = -8`
and `x = 8.`

This means that this process undergoes 8 alpha or `alpha` decays which means 8nuclei of He have been emitted

Alpha decays

`238U rarr 234Th`

`234Th rarr 230Th`

`230Th rarr 226Ra`

`226Ra rarr 222Rn`

`222Rn rarr 218Po`

`218Po rarr 214Pb`

`214Pb rarr 210Pb`

`210Pb rarr 206Pb`

Look at this image and you would see there are some beta decays too but that doesnt interfere the equation since beta decay doesnt change the mass but the number of protons.And in the equation we have not used the number of protons but the masses of the elements which are not changed. If we did the equation with number of protons the answer would be surely wrong.