Calculate de-Broglie wavelength associated with thermal neutrons at 27oc.
The de Broglie wavelength (λ) is given by the formula :
\[ \lambda = \frac{h}{p} \]
where :
\( \lambda \) = de Broglie wavelength,
\( h \) = Planck's constant (\(6.62607015 \times 10^{-34}\) J·s),
\( p \) = momentum of the particle.
For a particle with mass \( m \) and velocity \( v \), the momentum \( p \) can be calculated as :
\[ p = m \cdot v \]
For neutrons, the mass (\( m \)) is approximately \( 1.67493 \times 10^{-27}\) kg.
Now, let's calculate the velocity of a thermal neutron at 27 degrees Celsius. At thermal equilibrium, the average kinetic energy (\( \frac{1}{2} m v^2 \)) of the particle is equal to the thermal energy \( kT \), where \( k \) is the Boltzmann constant (\(1.380649 \times 10^{-23}\) J/K) and \( T \) is the temperature in Kelvin.
\[ \frac{1}{2} m v^2 = kT \]
Let's convert the temperature to Kelvin :
\[ T_{\text{Kelvin}} = T_{\text{Celsius}} + 273.15 \]
Now, we can rearrange the equation to solve for \( v \) :
\[ v = \sqrt{\frac{2kT}{m}} \]
Let's plug in the values and calculate \( v \) :
\[ T_{\text{Kelvin}} = 27 + 273.15 = 300.15 \text{ K} \]
\[ v = \sqrt{\frac{2 \cdot 1.380649 \times 10^{-23} \text{ J/K} \cdot 300.15 \text{ K}}{1.67493 \times 10^{-27} \text{ kg}}} \]
\[ v \approx 2186.83 \text{ m/s} \]
Now, we can calculate the de Broglie wavelength (\( \lambda \)):
\[ \lambda = \frac{h}{p} = \frac{6.62607015 \times 10^{-34} \text{ J·s}}{1.67493 \times 10^{-27} \text{ kg} \cdot 2186.83 \text{ m/s}} \]
\[ \lambda \approx 1.200 \times 10^{-10} \text{ meters} \]
So, the de Broglie wavelength associated with thermal neutrons at 27 degrees Celsius is approximately \( 1.200 \times 10^{-10} \) meters or \( 0.1200 \) nanometers.
Please note that the above calculation assumes non-relativistic velocities for the neutrons. At very high velocities, relativistic effects would need to be considered, but at thermal energies, the non-relativistic approximation is suitable.
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