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Question: Three \(\alpha -\)particles and one \(\beta -\)particle decaying takes place in series from an isoto...

Three α−\alpha -particles and one β−\beta -particle decaying takes place in series from an isotope 88Ra23888Ra^{238}.

Finally the isotope obtained will be.

A

84X22084X^{220}

B

86X22286X^{222}

C

83X22483X^{224}

D

83X21583X^{215}

Answer

83X22483X^{224}

Explanation

Solution

By using nα=A−A′4n_{\alpha} = \frac{A - A'}{4} and nβ=2nα−Z+Z′n_{\beta} = 2n_{\alpha} - Z + Z'

⇒ A′=A−4nα=236−4×3=224A' = A - 4n_{\alpha} = 236 - 4 \times 3 = 224

and Z′=(nβ−2nα+Z)=(1−2×3+88)=83Z' = (n_{\beta} - 2n_{\alpha} + Z) = (1 - 2 \times 3 + 88) = 83