A radioactive nuclide ""_86^226 Ra decays by emission of two alpha particles, one beta particle and gamma rays. Which of the following is the resulting nuclide X?

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KollegeApply · Accepted Answer

Initially element is represented by 86Ra226 a particle is represented by 2He4 this means it contains 2 protons and 2 neutrons so mass of a particle is 4 unit. Thus after emitting a particle mass number of element would decrease by 4 unit and atomic number of element would decrease by 2. So after emitting 2 a particle 86Ra226 would become 86 - 2 - 2Ra 226 - 4 - 4 = 82Ra218. $\beta$ particle is represented by -1e° after emitting a $\beta$ particle atomic number of element would increase by 1 unit and mass number of element would remain unchanged. So after emitting a p particle 82Ra218 would become 82 + 1Ra218 =83Ra218 Gamma radiations have zero mass and no charge so after emission of gamma radiation mass and atomic number remains unaltered. The daughter nucleus is represented by b x a b x a = 83Ra218 So answer is B.

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