When the radioactive isotope `._(88)Ra^(228)` decays in series by the emission of `3 alpha` and `1 beta` particle, the isotope finally formed is

When a radioactive isotope _88Ra^228 decays in series by the emission of 3 alpha particles and a beta particle, the isotope finally formed is

When a radio active isotopes "_88Ra^228 decays in series by the emission of three alpha -particles and a beta -particle, the isotopes finally formed is ...

When the radioactive isotope _(88)Ra^(226) decays in a series by emission of three aplha (alpha) and a beta (beta) particle, the isotope X which remains undecay is

When the radioactive isotope _(88)Ra^(226) decays in a series by emission of three aplha (alpha) and a beta (beta) particle, the isotope X which remains undecay is

A radioactive isotope _ZX^A decays in series by emitting 3 alpha -particles and 2 beta -particles. The resultant isotope will be :

The radioactive series to which ""_(88)^(228) Ra belongs is

""_(88)""^(224)Ra decays by a series emmision of three beta- particles and two alpha- particles. The end product of X is

In a radioactive disintegration a nucleus emits an alpha -particle first and then two beta -particles. Show that the final nucleus thus formed is the isotope of the first.

""_86^222Rn goes through radioactive distintegrations by successive emissions of alpha, alpha, beta, beta , alpha , beta " and " beta particles. Then final nucleus is