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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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a

$α = 3, β = 3$

b

$α = 6, β = 4$

c

$α = 6, β = 0$

d

$α = 1, β = 6$

answer is B.

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Detailed Solution

$m a s s n o d e c r e a s e s (232 - 208) = 24 s o 24 / 4 = 6 6 α p a r t i c l e s w i l l b e e m i t t e d t h e n a t o m i c n o d e c r e a s e s b y 6 x 2 = 12 b u t a c t u a l a t o m i c n o d e c r e a s e s b y (90 - 82) = 8 s o 12 - 8 = 4 β p a r t i c l e w i l l b e e m i t t e d$

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