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Question
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?
Options
223 x 83
218 x 83
218 x 84
217 x 85
218 x 85
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Solution
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.
β particle is represented by -1e° after emitting a β 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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