Question

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

Answer 1

Hint: De Broglie reasoned that matter, like light, may exhibit wave-particle duality, because light can behave both as a wave and as a particle (it contains packets of energy $ hv $ ).
The de Broglie waves exist as a closed-loop in the case of electrons travelling in circles around the nuclei in atoms, thus they can only exist as standing waves and fit evenly around the loop. As a result of this need, atoms' electrons orbit the nucleus in specific configurations, or states, known as stationary orbits.
 $ {{\lambda }_{neutron}}\alpha \dfrac{1}{\sqrt{T}} $ .

Complete answer:
The wavelength of a periodic wave is its spatial period, or the distance over which the wave's form repeats. It's the distance between two adjacent corresponding points of the same phase on the wave, such two adjacent crests, troughs, or zero crossings, and it's a feature of both travelling and standing waves, as well as other spatial wave patterns.
The square root of the mean of squares of the velocity of individual gas molecules is the root mean square velocity (RMS value). The arithmetic mean of the velocities of various molecules in a gas at a particular temperature is called average velocity.
Root mean square speed is given as $ \mathrm{v}_{\mathrm{rms}}=\sqrt{\dfrac{3 \mathrm{kT}}{\mathrm{m}}} $
Most probable speed is given as $ \mathrm{v}_{\mathrm{mp}}=\sqrt{\dfrac{2 \mathrm{kT}}{\mathrm{m}}} $
Average speed is given as $ \mathrm{v}_{\mathrm{av}}=\sqrt{\dfrac{8 \mathrm{kT}}{\pi \mathrm{m}}} $
From the above concepts it is understood $ \mathbf{v}_{\text {rms }} $ of gas molecule $ \propto \sqrt{T} $
According to the de-Broglie equation $ \lambda =\dfrac{\text{h}}{\text{mv}}=\dfrac{\text{h}}{\text{m}\times \text{K}\sqrt{\text{T}}} $ where K is a constant
So, $ \lambda \propto \dfrac{1}{\sqrt{\mathrm{T}}} $
 $ \lambda_{\text {neutron }} \alpha \dfrac{1}{\sqrt{T}} $
Here $ {{T}_{1}}={{27}^{o}}C=27+273=300K $
 $ {{T}_{2}}={{927}^{o}}C=927+273=1200K $
Upon substituting
 $ \Rightarrow \dfrac{\lambda_{1}}{\lambda_{2}}=\sqrt{\dfrac{T_{2}}{T_{1}}} $
 $ \Rightarrow \dfrac{\lambda_{1}}{\lambda_{2}}=\sqrt{\dfrac{273+927}{273+27}} $
 $ \Rightarrow \dfrac{\lambda_{1}}{\lambda_{2}}=\sqrt{\dfrac{1200}{300}} $
 $ \Rightarrow \dfrac{\lambda_{1}}{\lambda_{2}}=2 $
 $ {{\lambda }_{2}}=\dfrac{{{\lambda }_{1}}}{2} $ .

Note:
A wave's wavelength is determined by the medium through which it travels (for example, vacuum, air, or water). Sound waves, light waves, water waves, and periodic electrical impulses in a conductor are all examples of waves. A sound wave is a change in air pressure, whereas the intensity of the electric and magnetic fields varies in light and other electromagnetic radiation. Variations in the height of a body of water are known as water waves. Atomic locations change in a crystal lattice vibration.