The isotope `._(92)^(235)U` decays in a number of steps to an isotope of `._(82)^(207)Pb`. The groups of particle emitted in this process will be:

The isotope ""_(92)^(235) U decays in a number of steps to an isotope of ""_(87)^(207)Pb . The no of particles emitted in this process will be :

._(92)U^(235) undergoes successive disintegrations with the end product of ._(82)P^(203) . The number of alphaandbeta particles emitted are

The isotope of ""_(92)U^(238) decays by alpha -emission to an isotope of thorium (Th). The thorium isotope decays by beta -emission to an isotope of protactinium (Pa). Write down the equation to represents these two nuclear charges.

After losing a number of alpha and brta -particles, U_92^238 changed to Pb_82^206 . The total number of particles lost in this process is

In the given radioactive disintegration series ._(92)^(235)U rarr_(82)^(207)Pb Calculate number of alpha and number of beta particles emitted in this series.

In the given radioactive disintegration series ._(92)^(235)U rarr_(82)^(207)Pb Calculate difference between number of alpha and number of beta particles emitted in this series.

In the given radioactive disintegration series ._(92)^(235)U rarr_(82)^(207)Pb Calculate difference between number of alpha and number of beta particles emitted in this series.