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The de-broglie wavelength of neutron at ...

The de-broglie wavelength of neutron at `27^(@)C` is `lambda` . The wavelength at `927^(@)C` will be

A

`(lambda)/(9)`

B

`(lambda)/(4)`

C

`(lambda)/(2)`

D

`(lambda)/(3)`

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To solve the problem of finding the de Broglie wavelength of a neutron at 927°C given its wavelength at 27°C is λ, we will follow these steps: ### Step-by-Step Solution: 1. **Understanding de Broglie Wavelength**: The de Broglie wavelength (λ) is given by the formula: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. 2. **Relating Momentum to Kinetic Energy**: The momentum \( p \) can be expressed in terms of kinetic energy (KE): \[ p = \sqrt{2m \cdot KE} \] where \( m \) is the mass of the neutron. 3. **Kinetic Energy of an Ideal Gas**: For an ideal gas, the average kinetic energy can be expressed as: \[ KE = \frac{3}{2} kT \] where \( k \) is the Boltzmann constant and \( T \) is the absolute temperature in Kelvin. 4. **Substituting Kinetic Energy into the Wavelength Formula**: Substituting the expression for kinetic energy into the wavelength equation gives: \[ \lambda = \frac{h}{\sqrt{2m \cdot \frac{3}{2} kT}} = \frac{h}{\sqrt{3mkT}} \] 5. **Proportionality to Temperature**: From the above equation, we can see that: \[ \lambda \propto \frac{1}{\sqrt{T}} \] This means that the wavelength is inversely proportional to the square root of the temperature. 6. **Calculating Temperatures in Kelvin**: - For 27°C: \[ T_1 = 27 + 273 = 300 \, K \] - For 927°C: \[ T_2 = 927 + 273 = 1200 \, K \] 7. **Setting Up the Ratio of Wavelengths**: Using the proportionality: \[ \frac{\lambda_1}{\lambda_2} = \frac{\sqrt{T_2}}{\sqrt{T_1}} \] Substituting the temperatures: \[ \frac{\lambda}{\lambda_2} = \frac{\sqrt{1200}}{\sqrt{300}} \] 8. **Simplifying the Ratio**: \[ \frac{\lambda}{\lambda_2} = \frac{\sqrt{1200}}{\sqrt{300}} = \frac{\sqrt{1200/300}}{1} = \frac{\sqrt{4}}{1} = 2 \] Thus, we have: \[ \lambda_2 = \frac{\lambda}{2} \] ### Final Result: The wavelength at 927°C is: \[ \lambda_2 = \frac{\lambda}{2} \]

To solve the problem of finding the de Broglie wavelength of a neutron at 927°C given its wavelength at 27°C is λ, we will follow these steps: ### Step-by-Step Solution: 1. **Understanding de Broglie Wavelength**: The de Broglie wavelength (λ) is given by the formula: \[ \lambda = \frac{h}{p} ...
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