Question

(ii) An element as decays to 233R after emitting two alpha particles and one beta partic
Find the atomic number and atomic mass of the element S.​

Answer 1

Answer:

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−8R A−8 + −10β

After decay, given element is

85

R 222 . Therefore,A−8=222⇒A=230

(Atomic Mass)

Z−3=85⇒Z=88

Explanation:

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