Question

A nucleus with Z = 92 emits the following in a sequence:

α, β‾, β‾, α, α, α, α, α, β‾, β‾, α, β⁺, β⁺, α  

Then Z of the resulting nucleus is ______.

Options

• 76

• 78

• 82

• 74

Answer 1

A nucleus with Z = 92 emits the following in a sequence:

α, β‾, β‾, α, α, α, α, α, β‾, β‾, α, β⁺, β⁺, α  

Then Z of the resulting nucleus is 78.

Explanation:

The number of α-particles released = 8

Decrease in atomic number = 8 × 2 = 16

The number of β‾ particles released = 4

Increase in atomic number = 4 × 1 = 4

Also the number of β⁺ particles released is 2, which should decrease the atomic number by 2.

Therefore the final atomic number of resulting nucleus

= Z - 16 + 4 - 2

= Z - 14

= 92 - 14

= 78