Question

Radioactive disintegration of ${}_{88}^{226}Ra$ takes place in the following manner into $RaC$

$\mathbf{Ra}\overset{\mathbf{\quad -}\mathbf{\alpha}\mathbf{\quad}}{\rightarrow}\mathbf{Rn}\overset{\mathbf{\quad -}\mathbf{\alpha}\mathbf{\quad}}{\rightarrow}\mathbf{RaA}\overset{\mathbf{\quad -}\mathbf{\alpha}\mathbf{\quad}}{\rightarrow}\mathbf{RaB}\overset{\mathbf{\quad -}\mathbf{\beta}\mathbf{\quad}}{\rightarrow}\mathbf{RaC}$, Determine mass number and atomic number of RaC.

• A. 214 and 84
• B. 214 and 86
• C. 214 and 83
• D. 214 and 85

Answer 1

Answer: (C)

214 and 83

Answer 2

Parent element is ${}_{88}^{226}Ra$

Atomic mass = 226

Atomic number = 88

RaC is formed after the emission of 3 alpha particles. Mass of 3 alpha particles $= 3 \times 4 = 12$

So Atomic mass of RaC $= (226 - 12) = 214$

With emission of one α-particle, atomic number is decreased by 2 and with emission of β-particle, atomic number is increased by 1.

So Atomic number of RaC $= 88 - (3 \times 2) + 1 = 83$