Question

Radio active nucleus a finally transforms into a stable nucleus b then a and b can be

Answer 1

Answer:

8 alpha decays which release 8 alpha particles

Explanation:

U238→Pb206

(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 α decays which means 8nuclei of He have been emitted

Alpha decays

238U→234Th

234Th→230Th

230Th→226Ra

226Ra→222Rn

222Rn→218Po

218Po→214Pb

214Pb→210Pb

210Pb→206Pb

wikipedia

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.

I hope it will help you dear friend...