Question 276: Dual Nature of Matter and Radiation

Question

The de-Broglie wavelength of a neutron at $27^{\circ} \mathrm{C}$ is ' $\lambda_0$ '. What will be its wavelength at $927^{\circ} \mathrm{C}$ ?

Accepted Answer

Step 1: Recall de Broglie wavelength and thermal relation
The de Broglie wavelength:
$ \lambda = \frac{h}{p} $
For a particle in thermal equilibrium, its average kinetic energy is proportional to temperature:
$ \frac{1}{2} m v^2 \sim k_B T $
Thus,
$ v \propto \sqrt{T}, \quad p = m v \propto \sqrt{T}. $
So:
$ \lambda \propto \frac{1}{\sqrt{T}}. $
Step 2: Convert temperatures to Kelvin
At $27^\circ \mathrm{C}$: $T_1 = 27 + 273 = 300\ 	ext{K}$.
At $927^\circ \mathrm{C}$: $T_2 = 927 + 273 = 1200\ 	ext{K}$.
Step 3: Ratio of wavelengths
$ \frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{	frac{1}{4}} = 	frac{1}{2}. $
So:
$ \lambda_2 = \frac{\lambda_0}{2}. $
Ã¢ÂÂ Final Answer:
Option C: $\frac{\lambda_0}{2}$

Answer

$\frac{\lambda_0}{4}$

Answer

$\frac{\lambda_0}{3}$

Answer

$\frac{\lambda_0}{2}$

Answer

$\frac{3 \lambda_0}{2}$

Dual Nature of Matter and Radiation

The de-Broglie wavelength of a neutron at 27^{\circ} \mathrm{C} is ' \lambda_0 '. What will be its wavelength at 927^{\circ} \mathrm{C} ?

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