A nucleus at rest undergoes a decay emitting an a particle of de - Broglie wavelength ` lambda = 5.76 xx 10^(-15)m ` if the mass of the daughter nucleus is 223.610 amu and that of alpha particle is `4.002amu` , determine the total kinetic energy in the final state Hence , obtain the mass of the parent nucleus in amu (1 amu = 931.470 `MeV//e^(2)`)

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the momentum of the alpha particle using its de Broglie wavelength.
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{J s} \)) and \( p \) is the momentum.

Rearranging the formula to find momentum:
\[ p = \frac{h}{\lambda} \]

Substituting the values:
\[ p = \frac{6.63 \times 10^{-34}}{5.76 \times 10^{-15}} = 1.15 \times 10^{-19} \, \text{kg m/s} \]

### Step 2: Use conservation of momentum to find the momentum of the daughter nucleus.
Since the initial momentum is zero (the nucleus is at rest), the final momentum must also be zero. Therefore, the momentum of the daughter nucleus (\( p_D \)) is equal in magnitude but opposite in direction to the momentum of the alpha particle (\( p_\alpha \)):
\[ p_D = -p_\alpha = -1.15 \times 10^{-19} \, \text{kg m/s} \]

### Step 3: Calculate the total kinetic energy of the system.
The total kinetic energy (\( KE \)) can be expressed as the sum of the kinetic energies of the alpha particle and the daughter nucleus:
\[ KE = KE_\alpha + KE_D \]

Using the relationship between momentum and kinetic energy:
\[ KE = \frac{p^2}{2m} \]

Thus, the total kinetic energy can be expressed as:
\[ KE = \frac{p_\alpha^2}{2m_\alpha} + \frac{p_D^2}{2m_D} \]

Since \( p_D = -p_\alpha \):
\[ KE = \frac{p_\alpha^2}{2m_\alpha} + \frac{p_\alpha^2}{2m_D} = p_\alpha^2 \left( \frac{1}{2m_\alpha} + \frac{1}{2m_D} \right) \]

### Step 4: Substitute the values for the masses.
The masses are given as:
- Mass of the alpha particle (\( m_\alpha \)) = 4.002 amu
- Mass of the daughter nucleus (\( m_D \)) = 223.610 amu

Convert these masses to kilograms using \( 1 \, \text{amu} = 1.67 \times 10^{-27} \, \text{kg} \):
\[ m_\alpha = 4.002 \times 1.67 \times 10^{-27} \, \text{kg} = 6.67 \times 10^{-27} \, \text{kg} \]
\[ m_D = 223.610 \times 1.67 \times 10^{-27} \, \text{kg} = 3.73 \times 10^{-25} \, \text{kg} \]

### Step 5: Calculate the total kinetic energy.
Now substitute \( p_\alpha \) and the masses into the kinetic energy equation:
\[ KE = \left(1.15 \times 10^{-19}\right)^2 \left( \frac{1}{2 \times 6.67 \times 10^{-27}} + \frac{1}{2 \times 3.73 \times 10^{-25}} \right) \]

Calculating the individual terms:
\[ KE = \left(1.32 \times 10^{-38}\right) \left( \frac{1}{1.334 \times 10^{-26}} + \frac{1}{7.46 \times 10^{-26}} \right) \]
\[ KE = \left(1.32 \times 10^{-38}\right) \left(0.75 \times 10^{26}\right) = 1.0 \times 10^{-12} \, \text{J} \]

### Step 6: Convert kinetic energy to MeV.
To convert Joules to MeV:
\[ KE = \frac{1.0 \times 10^{-12}}{1.6 \times 10^{-19}} \approx 6.25 \, \text{MeV} \]

### Step 7: Calculate the mass defect.
Using the kinetic energy to find the mass defect (\( \Delta m \)):
\[ \Delta m = \frac{KE}{931.470 \, \text{MeV/amu}} = \frac{6.25}{931.470} \approx 0.0067 \, \text{amu} \]

### Step 8: Calculate the mass of the parent nucleus.
The mass of the parent nucleus (\( M_P \)) is given by:
\[ M_P = m_D + m_\alpha + \Delta m \]
Substituting the values:
\[ M_P = 223.610 + 4.002 + 0.0067 \approx 227.6187 \, \text{amu} \]

### Final Results:
- Total kinetic energy: \( 6.25 \, \text{MeV} \)
- Mass of the parent nucleus: \( 227.62 \, \text{amu} \)