The nucleus of an atom of `._(92)Y^(235)` initially at rest decays by emitting an `alpha` particle. The binding energy per nucleon of parent and daughter nuclei are `7.8MeV` and `7.835MeV` respectively and that of `alpha` particles is `7.07MeV//"nucleon"`. Assuming the daughter nucleus to be formed in the unexcited state and neglecting its share of energy in the reaction, calculate speed of emitted alpha particle. Take mass of `alpha` particle to be `6.68xx10^(-27)kg`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the decay equation
The decay of the nucleus can be represented as:
\[ _{92}^{235}Y \rightarrow _{90}^{231}X + _{2}^{4}He + Q \]
Where \(Q\) is the energy released during the decay.
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