De Broglie wavelength of 0.05 eV thermal neutron is

Options

A. `1.3Å`
B. `2Å`
C. `5.4Å`
D. `8Å`

Answer

To find the De Broglie wavelength of a thermal neutron with a kinetic energy of 0.05 eV, we can follow these steps:

### Step 1: Convert Kinetic Energy from eV to Joules
The kinetic energy (KE) is given in electron volts (eV). We need to convert this value into joules (J) using the conversion factor:
1 eV = \(1.6 \times 10^{-19}\) J.

\[ KE = 0.05 \, \text{eV} = 0.05 \times 1.6 \times 10^{-19} \, \text{J} = 8.0 \times 10^{-20} \, \text{J} \]

### Step 2: Calculate the Momentum
The momentum (p) of a particle can be related to its kinetic energy using the formula:
\[ KE = \frac{p^2}{2m} \]
Rearranging for momentum gives:
\[ p = \sqrt{2m \cdot KE} \]
The mass (m) of a neutron is approximately \(1.67 \times 10^{-27}\) kg. Plugging in the values:
\[ p = \sqrt{2 \times (1.67 \times 10^{-27} \, \text{kg}) \times (8.0 \times 10^{-20} \, \text{J})} \]

### Step 3: Calculate the De Broglie Wavelength
The De Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant, approximately \(6.63 \times 10^{-34} \, \text{Js}\). Now we can substitute the value of momentum we calculated in the previous step into this equation.

### Step 4: Substitute and Solve
First, calculate the momentum:
\[ p = \sqrt{2 \times (1.67 \times 10^{-27}) \times (8.0 \times 10^{-20})} = \sqrt{2.672 \times 10^{-46}} \approx 5.17 \times 10^{-23} \, \text{kg m/s} \]

Now, substitute \(p\) into the De Broglie wavelength formula:
\[ \lambda = \frac{6.63 \times 10^{-34}}{5.17 \times 10^{-23}} \approx 1.28 \times 10^{-11} \, \text{m} = 1.28 \, \text{Å} \]

### Final Answer
Thus, the De Broglie wavelength of a thermal neutron with a kinetic energy of 0.05 eV is approximately:
\[ \lambda \approx 1.28 \, \text{Å} \]