JEE Main 2003Annual, Paper 1PhysicsMediumAtoms and Nuclei183× since 2002A nucleus with Z=92Z=92Z=92 emits the following in a sequence: α,β−,β−,α,α,α,α,α,β−,β−,α,β+,β+,α\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,\alpha ,\alpha ,\alpha ,\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,{\beta ^ + },{\beta ^ + },\alphaα,β−,β−,α,α,α,α,α,β−,β−,α,β+,β+,α Then ZZZ of the resulting nucleus isAOption A. 767676BOption B. 787878COption C. 828282DOption D. 747474CheckSkip · show solution
