Question

The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable 20782Pb nucleus. The decays proceed through a series of alpha-particle and beta-particle emissions.A.How many of alpha-particle emissions are involved in this series?B.How many of beta-particle emissions are involved in this series?

Answer 1

Answer: There are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series

Step-by-step explanation:

Alpha Decay: In this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. The mass number is reduced by 4 units and atomic number is reduced by 2 units.

Beta Decay : It is a type of decay process, in which a proton gets converted to neutron and an electron. This is also known as -decay. In this the mass number remains same but the atomic number is increased by 1.

In radioactive decay the sum of atomic number or mass number of reactants must be equal to the sum of atomic number or mass number of products .

`_(92)^(235)\textrm{U}\rightarrow _(82)^(207)\textrm{Pb}+X_2^4\alpha+Y_(-1)^0e`

Thus for mass number : 235 = 207+4X

4X= 28

X = 7

Thus for atomic number : 92 = 82+2X-Y

2X- Y = 10

2(7) - Y= 10

14-10 = Y

Y= 4

`_(92)^(235)\textrm{U}\rightarrow _(82)^(207)\textrm{Pb}+7_2^4\alpha+4_(-1)^0e`

Thus there are 7 alpha-particle emissions and 4 beta-particle emissions involved in this series