First slide

Radio activity

Question

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 :

Moderate

A

$α = 3, β = 3$
B

$α = 6, β = 4$
C

$α = 6, β = 0$
D

$α = 1, β = 6$

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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