nuclear equation for the decay of iridium−192 with β and γ emission

Answer 1

I can't write this right. We should have a subscript on the left which is the atomic number, in this case 77. We should have a superscript on the left which is the mass number, in this case 192 and I just can do that on the computer. So the first number you understand is the atomic number and the second number (which follows the symbol) is what counts as the supscript; i.e., 77Ir192. The b particle is -1e0 and a gamma is oeo. Then the subscripts must add to the same value on both sides and the mass number must add to the fame value on both sides. So
77Ir192 ==> yXw + -1e0 + 0e0 So you know that
77 = y + (-1) + 0 so y must be 78 and 192 = w + 0 + 0 so w must be 192 . Look on the periodic table to find the element X = 78. That is Pt so we write the full equation as
77Ir192 = 78Pt192 + -1e0 + 0e0 where 0e0 is the gamma and -1e0 is the beta.