ICSE Class 10 Physics 2016 Solved Paper
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Question : 23
Total: 73
An element Z S A 85 R 222 2 α 1 β S
Solution:
On emission of 2 alpha particles, mass number is reduced by 8 and atomic number by 4 .
z S A → z − 4 Q A − 8 + 2 2 He 4
Now, on emission of 1 beta particle, mass number remains same but atomic number increases by 1 .
z − 4 Q A − 8 → Z − 3 R A − 8 + − 1 0 β
After decay, given element is85 R 222
A − 8 = 222 ⇒ A = 230
(Atomic Mass)
Z − 3 = 85 ∘ ⇒ Z = 88
(Atomic No.)
Now, on emission of 1 beta particle, mass number remains same but atomic number increases by 1 .
After decay, given element is
(Atomic Mass)
(Atomic No.)
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