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BINAY JHA

· started a discussion

· 1 Months ago

Q NO. 103 (Question: 298775.0 )

I DID NOT GET. PLS EXPLAIN

Question:

A radioactive decay process is given 92U238 ⟶ 82Pb214. In the above process emitted number of α and β particle are:-

Options:
A)

6, 3

B)

6, 2

C)

7, 4

D)

7, 3

Solution:

Ans: (b) 6, 2

Afrin Sultana

· commented

· 1 Months ago

alpha- 1 helium emitted, beta- 1 electron added.. so check mass number first and divide by 4. we get number of alpha particles... for beta, subtract atomic number by 2*alpha particles, we get beta particles..
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