  Question

# An imaginary radioactive element decays to form elements X1,X2,X3,X4,X5 and X6, by ejecting two β-particles, followed by an α-particle, and again two (β-particles followed by an α -particle). What are the mass number and the atomic number of X6?

A

22792X6

B

22692X6

C

22797X6

D

22790X6

Solution

## The correct option is A 22792X6 The symbolic representation of the reaction is:  23592Xβ−particle−−−−−−→X1β−particle−−−−−−→X2α−particle−−−−−−→X3β−particle−−−−−−→X4β−particle−−−−−−→X5α−particle−−−−−−→X6 In complete reaction, the total number of emitted a-particles = 2.  ∴ Loss of the total mass number due to two α-particles = 4 + 4 = 8 ∴ Loss of atomic number due to two α -particles = 2 + 2 = 4 In a complete reaction, the total number of emitted β - particles = 4. ∴ Loss of mass number due to four β -particles = 0 Gain of atomic number due to four β−particles=1×4=4 ∴ Total loss of mass number = 8 Total loss or gain of atomic number = 4 - 4 = 0 ∴ Mass number of X6=(235−8)=227 Atomic number of X6=(92−0)=92  Suggest corrections   