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 given its mass and kinetic energy, we can follow these steps:

### Step 1: Convert Kinetic Energy from eV to Joules
The kinetic energy (KE) is given as 0.04 eV. We need to convert this to Joules using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ Joules}\).

\[ KE = 0.04 \text{ eV} \times 1.6 \times 10^{-19} \text{ J/eV} = 6.4 \times 10^{-21} \text{ J} \]

### Step 2: Use the de Broglie Wavelength Formula
The de Broglie wavelength (\(\lambda\)) can be calculated using the formula:

\[ \lambda = \frac{h}{p} \]

where \(h\) is Planck's constant and \(p\) is the momentum of the particle. The momentum can be expressed in terms of kinetic energy:

\[ p = \sqrt{2m \cdot KE} \]

### Step 3: Calculate Momentum
Substituting the values into the momentum formula:

- Mass of neutron, \(m = 1.6 \times 10^{-27} \text{ kg}\)
- Kinetic energy, \(KE = 6.4 \times 10^{-21} \text{ J}\)

\[ p = \sqrt{2 \cdot (1.6 \times 10^{-27} \text{ kg}) \cdot (6.4 \times 10^{-21} \text{ J})} \]

Calculating this gives:

\[ p = \sqrt{2 \cdot 1.6 \times 10^{-27} \cdot 6.4 \times 10^{-21}} = \sqrt{2.048 \times 10^{-46}} \approx 4.53 \times 10^{-23} \text{ kg m/s} \]

### Step 4: Calculate de Broglie Wavelength
Now substituting \(p\) into the de Broglie wavelength formula:

Planck's constant \(h = 6.626 \times 10^{-34} \text{ J s}\)

\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J s}}{4.53 \times 10^{-23} \text{ kg m/s}} \approx 1.46 \times 10^{-11} \text{ m} \]

### Step 5: Convert to Angstroms
Since \(1 \text{ angstrom} = 10^{-10} \text{ m}\):

\[ \lambda \approx 1.46 \times 10^{-11} \text{ m} = 0.146 \text{ angstroms} \]

### Final Answer
The de Broglie wavelength of the neutron is approximately \(1.46 \times 10^{-11} \text{ m}\) or \(0.146 \text{ angstroms}\).

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