When the radioactive `isotope _(88)Ra^(226)` decays in a series by emission of three aplha `(alpha)` and a beta `(beta)` particle, the isotope X which remains undecay is
A. `83^(X)^(214)`
B. `84^(X)^(218)`
C. `84^(X)^(220)`
D. `87^(X)^(223)`


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Correct Answer - A
Due to emission os `3alpha` particles, the mass number is reduced by 12 while atomic number is decresed by 6. Due to emission of `beta` particle the atomic number is increased by one while there is no change in mass number is decreased by 5. so isotope X is is `X^(214)`