A nuclear reaction involves the hypothetical element X, which has an atomic numb

Question

A nuclear reaction involves the hypothetical element X, which has an atomic number of 169, as a reactant. The 361X nuclide reacts with 6 neutrons yielding a nuclide of element Y along with, altogether, 4 beta particles, 4 neutrons, 5 alpha particles, 5 positrons, and 5 gamma rays. The number of neutrons in the nucleus of the resultant nuclide of Y would be? Please provide explanation!

Explanation / Answer

361X + 6n ---> Y + 4 (e-) + 4n + 5 (4He+) + (5e+) + 5 (gamma) positoron has atm no 1, mass no =0 beta particle has atomic no -1 and mass no =0 alpha particle has atm no = 2 , mass no =4 neutron has atm no = 0 , mass no =1 based on these 361 + 6 = mass no of Y + 4 + 5(4) mass no of Y = 343 169 = atm no of Y + 4(-1) + 5(2) + 5(1) atm no of Y = 158 so number of neutrons = mass no -atm no = 343-158 = 185

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