Atomic mass number of an element is 232 and its atomic number is 90. The end product, of this radioactive element is an isotope of lead (""_(82)^(208)Pb). The number of alpha and beta particles emitted is :

Atomic mass number of an element is 232 and its atomic number is 90. T

1.	Atomic mass number of an element is 232 and its atomic number is 90. The end product, of this radioactive element is an isotope of lead (""_(82)^(208)Pb). The number of alpha and beta particles emitted is :
Answer» `alpha =3 and BETA=3` `alpha=6 and beta=4` `alpha =6 and beta=0` `alpha =1 and beta=6` Solution :LET x be `alpha`- decays &y was `beta` decay. Then `4X=(232-208) & (2x-y)=90-82` `x=6""alpha=6` `y=4""beta=4`

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Atomic mass number of an element is 232 and its atomic number is 90. The end product, of this radioactive element is an isotope of lead (""_(82)^(208)Pb). The number of alpha and beta particles emitted is :

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