For what kinetic energy of a neutron will the associated de-Broglie wavelength be `1.40 xx 10^(-10)m` ? Mass of neutron = `1.675 xx 10^(-27) kg, h = 6.63 xx 10^(-34) Js`.

Answer

To find the kinetic energy of a neutron associated with a de-Broglie wavelength of \( \lambda = 1.40 \times 10^{-10} \, \text{m} \), we can follow these steps:

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

### Step 2: Rearrange to find momentum
Rearranging the formula to solve for momentum \( p \):
\[ p = \frac{h}{\lambda} \]

### Step 3: Substitute the values
Substituting the known values into the equation:
\[ p = \frac{6.63 \times 10^{-34} \, \text{Js}}{1.40 \times 10^{-10} \, \text{m}} \]

### Step 4: Calculate momentum
Calculating the above expression:
\[ p = \frac{6.63}{1.40} \times 10^{-34 + 10} = 4.7357 \times 10^{-24} \, \text{kg m/s} \]

### Step 5: Use the kinetic energy formula
The kinetic energy \( KE \) is given by the formula:
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the neutron (\( 1.675 \times 10^{-27} \, \text{kg} \)).

### Step 6: Substitute momentum into the kinetic energy formula
Substituting the value of \( p \) into the kinetic energy formula:
\[ KE = \frac{(4.7357 \times 10^{-24})^2}{2 \times 1.675 \times 10^{-27}} \]

### Step 7: Calculate \( p^2 \)
Calculating \( p^2 \):
\[ p^2 = (4.7357 \times 10^{-24})^2 = 22.42 \times 10^{-48} \, \text{kg}^2 \text{m}^2/\text{s}^2 \]

### Step 8: Substitute \( p^2 \) into the kinetic energy formula
Now substituting \( p^2 \) into the kinetic energy formula:
\[ KE = \frac{22.42 \times 10^{-48}}{2 \times 1.675 \times 10^{-27}} \]

### Step 9: Calculate the denominator
Calculating the denominator:
\[ 2 \times 1.675 \times 10^{-27} = 3.35 \times 10^{-27} \]

### Step 10: Calculate kinetic energy
Finally, calculating \( KE \):
\[ KE = \frac{22.42 \times 10^{-48}}{3.35 \times 10^{-27}} = 6.69 \times 10^{-21} \, \text{J} \]

### Final Answer
The kinetic energy of the neutron is:
\[ KE \approx 6.69 \times 10^{-21} \, \text{J} \]

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