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

Answer 1

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 - 2}Ra ^{226 - 4 - 4} = ₈₂Ra²¹⁸.
β particle is represented by -1^e° 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  + 1}Ra²¹⁸ =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.