If `._92 U^238` undergoes successively `8 alpha-`decays and `6 beta-`decays, then resulting nucleus is.

The isotope ._(92)U^(238) successively undergoes eight alpha -decays and six beta -decays. The resulting isotope is found to be ._(92-5y)X^(238-4x) . Find the value of ratio x//y .

A nucleus ._(90)^(229)X under goes alpha -decay and beta -decay and the resultant nucleus is ._(73)^(181)Y . Find number of beta -decay.

A nucleus ._(90)^(229)X under goes alpha -decay and beta -decay and the resultant nucleus is ._(73)^(181)Y . Find number of beta -decay.

._92U^238 decays by emitting successively 8 alpha -particles and 8 beta -particles. Determine the mass number and atomic number of the new element and express it in symbol.

._(92)U^(238) emits 8 alpha- particles and 6 beta- particles. The n//p ratio in the product nucleus is

._(92)U^(238) emits 8 alpha- particles and 6 beta- particles. The n//p ratio in the product nucleus is

The nucleus resulting from ""_(92)U^(238) after successive emission of two alpha and four beta -particles is