The sequence of decay of radioactive nucleus is `D overset alphato D_1oversetbeta toD_2 oversetalpha toD_3`. If nucleon number and atomic number of `D_2` are 176 and 71 respectively, what are their values for `D` and `D_3`?

Explanation

As mass number of each `alpha` particles is `4` units and its charge number is `2` units, therefore, for `D_(4)A=176-8=168 Z=71-4=67`
Now charge no. of `beta` is `-1` and its mass number is zero, therefore, for `D`
`A=176+0+4=180 Z=71-1+2=72`