`._^(239)Pu._(94)` is undergoing `alpha-decay` according to the equation `._(94)^(239)Pu rarr (._(97)^(235)U) +._2^4 He` . The energy released in the process is mostly kinetic energy of the `alpha`-particle. However, a part of the energy is released as `gamma` rays. What is the speed of the emiited `alpha`-particle if the `gamma` rays radiated out have energy of `0.90 MeV`? Given: Mass of `._(94)^(239)Pu =239.05122 u`, mass of `(._(97)^(235)U)=235.04299 u` and mass of `._1^4He =4.002602 u (1u =931 MeV)`.

Explanation

To find the speed of the emitted alpha particle during the alpha decay of Plutonium-239, we can follow these steps:

### Step 1: Calculate the mass defect (Δm)
The mass defect is the difference between the mass of the parent nucleus and the total mass of the products.

Given:
- Mass of \( _{94}^{239}Pu = 239.05122 \, u \)
- Mass of \( _{97}^{235}U = 235.04299 \, u \)
- Mass of \( _{2}^{4}He = 4.002602 \, u \)

Using the equation:
\[ \Delta m = \text{Mass of parent} - (\text{Mass of daughter} + \text{Mass of alpha particle}) \]
Substituting the values:
\[ \Delta m = 239.05122 \, u - (235.04299 \, u + 4.002602 \, u) \]
\[ \Delta m = 239.05122 \, u - 239.045592 \, u = 0.005628 \, u \]

### Step 2: Convert mass defect to energy released (E)
Using the mass-energy equivalence \( E = \Delta m \cdot c^2 \), where \( 1 \, u = 931 \, MeV \):
\[ E = 0.005628 \, u \times 931 \, MeV/u = 5.24 \, MeV \]

### Step 3: Determine the kinetic energy of the alpha particle
The total energy released is partly in the form of gamma rays. Given that the energy of gamma rays is \( 0.90 \, MeV \), the kinetic energy (KE) of the alpha particle can be calculated as:
\[ KE_{\alpha} = E - E_{\gamma} \]
\[ KE_{\alpha} = 5.24 \, MeV - 0.90 \, MeV = 4.34 \, MeV \]

### Step 4: Convert kinetic energy to joules
To convert MeV to joules, use the conversion \( 1 \, MeV = 1.6 \times 10^{-13} \, J \):
\[ KE_{\alpha} = 4.34 \, MeV \times 1.6 \times 10^{-13} \, J/MeV = 6.944 \times 10^{-13} \, J \]

### Step 5: Calculate the speed of the alpha particle
The kinetic energy is also given by the equation:
\[ KE = \frac{1}{2} mv^2 \]
Where \( m \) is the mass of the alpha particle. The mass of the alpha particle is \( 4.002602 \, u \):
\[ m = 4.002602 \, u \times 1.66 \times 10^{-27} \, kg/u = 6.645 \times 10^{-27} \, kg \]

Now, rearranging the kinetic energy formula to solve for \( v \):
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \cdot 6.944 \times 10^{-13} \, J}{6.645 \times 10^{-27} \, kg}} = \sqrt{2.09 \times 10^{14}} \approx 1.45 \times 10^7 \, m/s \]

### Final Answer:
The speed of the emitted alpha particle is approximately \( 1.45 \times 10^7 \, m/s \).

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