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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 ${}_{82}^{208}P b$ .The number of alpha and beta particles emitted is :

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α =3, β = 3

α =6, β =4

α =6, β =0

α =1, β = 6

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detailed solution

Correct option is B

mass no decreases 232-208= 24 so 24/4= 6 6α particles will be emitted then atomic no decreases by 6x2 =12 but actual atomic no decreases by 90-82=8 so 12-8 = 4 β particle will be emitted

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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 P 82208b .The number of alpha and beta particles emitted is :

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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 ${}_{82}^{208}P b$ .The number of alpha and beta particles emitted is :