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?
The symbolic representation of the reaction is:
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