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

Mar 20, 2017

$\text{8 alpha decays which release 8 alpha particles}$

#### Explanation:

${U}^{238} \rightarrow P {b}^{206}$

(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

$206 = 238 - \left(4 x\right)$

Solve for x

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

$- 4 x = - 32$

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

$238 U \rightarrow 234 T h$

$234 T h \rightarrow 230 T h$

$230 T h \rightarrow 226 R a$

$226 R a \rightarrow 222 R n$

$222 R n \rightarrow 218 P o$

$218 P o \rightarrow 214 P b$

$214 P b \rightarrow 210 P b$

$210 P b \rightarrow 206 P b$ 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.