The naturally occurring radioactive decay series that begins with 23592U stops with formation of the stable...

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.

Part A

How many of alpha-particle emissions are involved in this series?

Part B

How many of beta-particle emissions are involved in this series?

Expert Solution

Ans:-Part A . 7 alpha- particles (₂He⁴) and

Part B .4 beta -particles (_-1e⁰)

Explanation:-

alpha-particle emisson :- Alpha-particle is ₂He⁴ , here 2 is the atomic number and 4 is the mass number of He . therefore in alpha-particle emission atomic number is increased by 2 units while mass number is increased by 4 units .

beta - particle emission:- Beta-particle is _-1e⁰ , here -1 is the atomic number and 0 is the mass number . therefore in beta decay atomic number is decreased by 1 unit while there is no change in mass number.

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 .

Therefore overall equation of decay will be :

₉₂U²³⁵ ----------------> ₈₂Pb²⁰⁷ + 7 ₂He⁴ + 4 _-1e⁰

Sum of atomic number of reactants (92) = Sum of atomic number of products (82 + 14 - 4= 92 ) and also

Sum of mass number of reactants ( 235 ) = Sum of mass number of products (207 + 28 + 0 = 235)

queen_honey_blossom answered 2 years ago

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