Energy of a particle at absolute temperature T is of the order kT. Calculate the wavelength of thermal neutrons at `27^(@)C`. Take mass of neutron = `1.67 xx 10^(-27)kg, h =6.6 xx 10^(-34)Js` and k = 1.38 J `" mol"^(-1)K^(-1)`.

Answer

To calculate the wavelength of thermal neutrons at a temperature of 27°C, we will follow these steps:

### Step 1: Convert the temperature from Celsius to Kelvin
The absolute temperature \( T \) in Kelvin is given by:
\[ T = 27 + 273 = 300 \, \text{K} \]

### Step 2: Calculate the thermal energy
The energy of a particle at absolute temperature \( T \) is given by:
\[ E = kT \]
where \( k \) is the Boltzmann constant. Given \( k = 1.38 \times 10^{-23} \, \text{J/K} \), we can calculate the thermal energy:
\[ E = kT = (1.38 \times 10^{-23} \, \text{J/K}) \times (300 \, \text{K}) = 4.14 \times 10^{-21} \, \text{J} \]

### Step 3: Relate thermal energy to kinetic energy
The kinetic energy of the neutron can be expressed as:
\[ E = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the neutron. Rearranging this gives:
\[ v^2 = \frac{2E}{m} \]
Substituting the values, where the mass of the neutron \( m = 1.67 \times 10^{-27} \, \text{kg} \):
\[ v^2 = \frac{2 \times 4.14 \times 10^{-21} \, \text{J}}{1.67 \times 10^{-27} \, \text{kg}} = \frac{8.28 \times 10^{-21}}{1.67 \times 10^{-27}} \approx 4.95 \times 10^{6} \]
Taking the square root to find \( v \):
\[ v \approx \sqrt{4.95 \times 10^{6}} \approx 2224 \, \text{m/s} \]

### Step 4: Calculate the de Broglie wavelength
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum, given by \( p = mv \). Thus:
\[ p = m \cdot v = (1.67 \times 10^{-27} \, \text{kg}) \cdot (2224 \, \text{m/s}) \approx 3.71 \times 10^{-24} \, \text{kg m/s} \]
Now substituting into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34} \, \text{Js}}{3.71 \times 10^{-24} \, \text{kg m/s}} \approx 1.79 \times 10^{-10} \, \text{m} = 1.79 \, \text{Å} \]

### Final Answer
The wavelength of thermal neutrons at 27°C is approximately:
\[ \lambda \approx 1.79 \, \text{Å} \]

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