Find the speed of particle produced during the decay of `""_(92)p^(235) to ""_(90)q^(231)` . Assume that the energy shared by the daughter nucleus is negligible. B.E./nucleon of`""_(92)p^(235) =7.81 MeV`.
B.E./nucleon of `""_(90)Q^(231) = 7.834 MeV`
B.E./nucleon of `alpha`-particle = 7.02 MeV
Mass of `alpha`-particle `=6.7 xx 10^(-27)` kg

Explanation

To find the speed of the alpha particle produced during the decay of \( _{92}P^{235} \) to \( _{90}Q^{231} \), we can follow these steps:

### Step 1: Write the decay reaction
The decay can be represented as:
\[ _{92}P^{235} \rightarrow _{90}Q^{231} + _{2}\alpha^{4} \]
This shows that the parent nucleus \( P^{235} \) decays into the daughter nucleus \( Q^{231} \) and an alpha particle.
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