A radioactive nucleus ${}${92}^{238}\text{U} continuously decays and changes into ${}{82}^{206}\text{Pb}$. In this decay, how many α and β particles are emitted ?

To determine the number of α and β decays, compare the changes in mass number and atomic number.

• Parent nucleus : ${}_{92}^{238}\text{U}$
• Daughter nucleus : ${}_{82}^{206}\text{Pb}$

Change in mass number = 238 - 206 = 32

And

Change in atomic number = 92 - 82 = 10

Since each α-particle reduces the mass number by 4.

Then

Number of α particles = $\dfrac{32}{4}$ = 8

As each α decay reduces atomic number by 2 then,

Change in atomic number of parent nucleus = 92 - (8 x 2) = = 92 - 16 = 76

But final atomic number is 82, so we need an increase:

82 − 76 = 6

Since each β decay increases atomic number by 1, so :

Number of β particles emitted = 6

Hence, 8 α and 6 β particles are emitted.