A nucleus with Z =92 emits the following in a sequence:
`alpha,beta^(-),beta^(-),alpha,alpha,alpha,alpha,alpha,beta^(-),beta^(-),alpha,beta^(+),beta^(+),alpha`. The Z of the resulting nucleus is

A nucleus with Z=92 emits the following in a sequence alpha, alpha, beta^(-), beta^(-), alpha, alpha, alpha, alpha, beta^(-), beta^(-), alpha, beta^(+), alpha . The Z of the resulting nucleus is

A nucleus with z= 92 units emits the following in a sequence. alpha, beta^(-) , alpha, beta^(-) , alpha, alpha, alpha, alpha , beta^(-) , beta^(-) , alpha,beta^+ , beta^+ , alpha . The Z of the resulting nucleus is

A nucleus with atomic number Z (Z = 92) emits the following particles in a sequence alpha, beta^(-), alpha, alpha, beta^(-), beta^(+), alpha, beta^(-), alpha, beta^(+) . The atomic number of the resulting nucleus is

If the following disintegrations happen to an element of Z= 92 : alpha,beta^-1,beta^-1, alpha,alpha,alpha,alpha,alpha,beta^-,beta^-,alpha,beta^+,beta^+,alpha . Then what will be the final Z?—

|(alpha,-beta,0),(0,alpha,beta),(beta,0,alpha)|=0 , then

| alpha alpha1 beta F|=(alpha-P)(beta-alpha)

If f(alpha,beta)=|(cos alpha,-sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)|, then

If f(alpha,beta)=|(cos alpha,-sin alpha,1),(sin alpha,cos alpha,1),(cos(alpha+beta),-sin(alpha+beta),1)|, then