Step 1: Understanding the Concept:
Thermal neutrons are particles in thermal equilibrium with their surroundings.
Their kinetic energy is directly proportional to the absolute temperature.
The de-Broglie wavelength associates a wave characteristic to these moving particles based on their momentum. Step 2: Key Formula or Approach:
The kinetic energy \( E \) of a gas particle (or thermal neutron) at absolute temperature \( T \) is:
\[ E = \frac{3}{2} k_B T \]
The de-Broglie wavelength \( \lambda \) is related to kinetic energy by:
\[ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE}} \]
Substituting the expression for \( E \):
\[ \lambda = \frac{h}{\sqrt{2m(\frac{3}{2}k_B T)}} = \frac{h}{\sqrt{3m k_B T}} \]
From this, we see that the wavelength is inversely proportional to the square root of the absolute temperature:
\[ \lambda \propto \frac{1}{\sqrt{T}} \]
Step 3: Detailed Explanation:
First, we must convert the given Celsius temperatures to Kelvin (absolute temperature).
Initial temperature \( T_1 = 27^\circ\text{C} = 27 + 273 = 300\text{ K} \).
Final temperature \( T_2 = 927^\circ\text{C} = 927 + 273 = 1200\text{ K} \).
Let the initial wavelength be \( \lambda_1 = \lambda_0 \) and the final wavelength be \( \lambda_2 \).
Using the proportionality relation \( \lambda \propto \frac{1}{\sqrt{T}} \), we can set up a ratio:
\[ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} \]
Substitute the values of the temperatures into the ratio:
\[ \frac{\lambda_2}{\lambda_0} = \sqrt{\frac{300}{1200}} \]
Simplify the fraction inside the square root:
\[ \frac{\lambda_2}{\lambda_0} = \sqrt{\frac{1}{4}} \]
Take the square root:
\[ \frac{\lambda_2}{\lambda_0} = \frac{1}{2} \]
Now, solve for the final wavelength \( \lambda_2 \):
\[ \lambda_2 = \frac{\lambda_0}{2} \]
Step 4: Final Answer:
The wavelength at \( 927^\circ\text{C} \) will be \( \frac{\lambda_0}{2} \).