For what kinetic energy of a neutron will the associated de broglie wavelenght be `1.04 xx 10^(-10) m` ?

Options

A. `6,29 xx 10 ^(-23)j`
B. `5.99 xx 10^(-19j`
C. `6.69 xx 10^(-21)j`
D. `9.66 xx 10^(-21)j`

Explanation

To find the kinetic energy of a neutron for which the associated de Broglie wavelength is \( \lambda = 1.04 \times 10^{-10} \, \text{m} \), we can follow these steps:

### Step 1: Write 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 and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
The momentum \( p \) of a particle can be expressed in terms of its kinetic energy \( KE \) as:
\[ p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the particle.

### Step 3: Substitute momentum in the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]

### Step 4: Rearrange the formula to solve for kinetic energy
Rearranging the equation to solve for kinetic energy \( KE \):
\[ KE = \frac{h^2}{2m \cdot \lambda^2} \]

### Step 5: Substitute known values
Now we can substitute the known values:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of a neutron \( m = 1.67 \times 10^{-27} \, \text{kg} \)
- Wavelength \( \lambda = 1.04 \times 10^{-10} \, \text{m} \)

Substituting these values into the equation:
\[ KE = \frac{(6.63 \times 10^{-34})^2}{2 \cdot (1.67 \times 10^{-27}) \cdot (1.04 \times 10^{-10})^2} \]

### Step 6: Calculate the kinetic energy
Calculating the numerator:
\[ (6.63 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{J}^2 \]

Calculating the denominator:
\[ 2 \cdot (1.67 \times 10^{-27}) \cdot (1.04 \times 10^{-10})^2 = 2 \cdot (1.67 \times 10^{-27}) \cdot (1.0816 \times 10^{-20}) = 3.60 \times 10^{-47} \, \text{kg m}^2 \]

Now substituting these values back into the equation for kinetic energy:
\[ KE = \frac{4.39 \times 10^{-67}}{3.60 \times 10^{-47}} \approx 1.22 \times 10^{-20} \, \text{J} \]

### Final Answer
The kinetic energy of the neutron is approximately:
\[ KE \approx 1.22 \times 10^{-20} \, \text{J} \]