Calculate de Broglie wavelenth of a neutron (mass, `m=1.6xx10^(-27)` kg) moving with kinetic energy of 0.04 eV-

Options

A. 146 `"Å"`
B. 14.6 `"Å"`
C. 1460 `"Å"`
D. 1.46 `"Å"`

Explanation

To calculate the de Broglie wavelength of a neutron with a given mass and kinetic energy, we can follow these steps:

### Step 1: Write down the formula for de Broglie wavelength
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2 m KE}} \]
where:
- \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the particle (in kg),
- \( KE \) is the kinetic energy (in joules).

### Step 2: Convert kinetic energy from electron volts to joules
The kinetic energy is given as \( 0.04 \, \text{eV} \). To convert this to joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ KE = 0.04 \, \text{eV} \times 1.6 \times 10^{-19} \, \text{J/eV} = 6.4 \times 10^{-21} \, \text{J} \]

### Step 3: Substitute the values into the de Broglie wavelength formula
Now we can substitute the values into the formula:
- Mass of neutron, \( m = 1.6 \times 10^{-27} \, \text{kg} \)
- Kinetic energy, \( KE = 6.4 \times 10^{-21} \, \text{J} \)

Substituting these values gives:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-27} \times 6.4 \times 10^{-21}}} \]

### Step 4: Calculate the denominator
First, calculate \( 2 \times 1.6 \times 10^{-27} \times 6.4 \times 10^{-21} \):
\[ 2 \times 1.6 \times 10^{-27} \times 6.4 \times 10^{-21} = 2.048 \times 10^{-47} \]
Now, take the square root:
\[ \sqrt{2.048 \times 10^{-47}} = 4.53 \times 10^{-24} \]

### Step 5: Calculate the de Broglie wavelength
Now substitute back into the equation for λ:
\[ \lambda = \frac{6.626 \times 10^{-34}}{4.53 \times 10^{-24}} \approx 1.46 \times 10^{-10} \, \text{m} \]

### Step 6: Convert to angstroms
Since \( 1 \, \text{angstrom} = 10^{-10} \, \text{m} \):
\[ \lambda \approx 1.46 \, \text{angstroms} \]

### Final Answer
The de Broglie wavelength of the neutron is approximately \( 1.46 \, \text{Å} \).

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