The de-Broglie wavelength of a neutron at 27^ C is ' ₀ '. What will be its wavelength at 927^ C ?

Step-by-step solution

The de-Broglie wavelength of a neutron is given by = h p , where h is Planck's constant and p is the momentum. For a particle in thermal equilibrium, the average kinetic energy relates to temperature through K = 3 2 kT , where k is Boltzmann's constant. Since kinetic energy and momentum are connected by K = p^2 2m , substitution yields = h 3mkT . The wavelength varies inversely with the square root of temperature: 1 T . For initial temperature T₁ = 300 , K and final temperature T₂ = 1200 , K , the wavelength ratio