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 dougther nuclei are 7.8MeV and 7.835MeV respectively and that of `alpha` particles is `7.07MeV//"nucleon"`. Assuming the dougther 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 step-by-step, we will follow these steps:

### Step 1: Write the decay reaction
The decay of the nucleus can be represented as:
\[ ^{235}_{92}Y \rightarrow ^{231}_{90}X + ^{4}_{2}\alpha + E \]

### Step 2: Calculate the energy released in the decay
The energy released (E) in the decay can be calculated using the binding energy per nucleon of the parent and daughter nuclei along with the alpha particle. The formula for energy released is:
\[ E = \left( \text{Binding Energy per nucleon of } X \times \text{Mass number of } X \right) + \left( \text{Binding Energy per nucleon of } \alpha \times \text{Mass number of } \alpha \right) - \left( \text{Binding Energy per nucleon of } Y \times \text{Mass number of } Y \right) \]

Given:
- Binding energy per nucleon of \(Y\) = 7.8 MeV
- Binding energy per nucleon of \(X\) = 7.835 MeV
- Binding energy per nucleon of \(\alpha\) = 7.07 MeV
- Mass number of \(Y\) = 235
- Mass number of \(X\) = 231
- Mass number of \(\alpha\) = 4

Substituting the values:
\[ E = (7.835 \times 231) + (7.07 \times 4) - (7.8 \times 235) \]

Calculating each term:
- \(7.835 \times 231 = 1809.885 \, \text{MeV}\)
- \(7.07 \times 4 = 28.28 \, \text{MeV}\)
- \(7.8 \times 235 = 1833.0 \, \text{MeV}\)

Now substituting these values back into the equation:
\[ E = 1809.885 + 28.28 - 1833.0 = 5.165 \, \text{MeV} \]

### Step 3: Convert energy from MeV to Joules
To convert MeV to Joules, we use the conversion factor \(1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}\):
\[ E = 5.165 \times 1.6 \times 10^{-13} = 8.264 \times 10^{-13} \, \text{J} \]

### Step 4: Relate the energy to the kinetic energy of the alpha particle
The kinetic energy (K.E.) of the emitted alpha particle can be expressed as:
\[ K.E. = \frac{1}{2} m_{\alpha} v^2 \]
Where \(m_{\alpha}\) is the mass of the alpha particle and \(v\) is its speed.

### Step 5: Solve for the speed of the alpha particle
Setting the kinetic energy equal to the energy released:
\[ \frac{1}{2} m_{\alpha} v^2 = E \]
Rearranging for \(v\):
\[ v = \sqrt{\frac{2E}{m_{\alpha}}} \]

Substituting the values:
- \(E = 8.264 \times 10^{-13} \, \text{J}\)
- \(m_{\alpha} = 6.68 \times 10^{-27} \, \text{kg}\)

Calculating \(v\):
\[ v = \sqrt{\frac{2 \times 8.264 \times 10^{-13}}{6.68 \times 10^{-27}}} \]
\[ v = \sqrt{\frac{1.6528 \times 10^{-12}}{6.68 \times 10^{-27}}} \]
\[ v = \sqrt{2.477 \times 10^{14}} \approx 1.577 \times 10^{7} \, \text{m/s} \]

### Final Answer
The speed of the emitted alpha particle is approximately:
\[ v \approx 1.577 \times 10^{7} \, \text{m/s} \]