A radioactive nucleus M undergoes series of decays given below:
\( M \overset{\beta^{-}\ \text{decay}}{\longrightarrow} N \overset{\gamma\ \text{decay}}{\longrightarrow} O \overset{\alpha\ \text{decay}}{\longrightarrow} P \)
The mass number of M is 150 and Atomic No of M is 64. The Atomic number of P is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 63
Each kind of decay changes the atomic number \( Z \) and the mass number \( A \) in a fixed way. In \( \beta^{-} \) decay a neutron turns into a proton, so \( Z \) rises by 1 and \( A \) is unchanged. In \( \gamma \) decay only energy is emitted, so neither changes. In \( \alpha \) decay a helium nucleus leaves, so \( Z \) falls by 2 and \( A \) falls by 4.
\( M:\ Z = 64,\ A = 150 \)
\( N\ (\beta^{-}):\ Z = 65,\ A = 150 \)
\( O\ (\gamma):\ Z = 65,\ A = 150 \)
\( P\ (\alpha):\ Z = 63,\ A = 146 \)
So the atomic number of P is 63.