The de-Broglie wavelength of a neutron at 927 °C is λ. What will be its wavelength at 27 °C?

The de-Broglie wavelength of a particle is given by the formula:

λ = h / p

where h is Planck's constant and p is the momentum of the particle. The momentum of a particle can be related to its temperature by the formula:

p = sqrt(2mk*T)

where m is the mass of the particle, k is Boltzmann's constant, and T is the temperature in Kelvin.

Therefore, the de-Broglie wavelength can be written in terms of the temperature as:

λ = h / sqrt(2mk*T)

Now, we are given that the de-Broglie wavelength of a neutron at 927 °C (or 1200 K) is λ. We want to find the wavelength at 27 °C (or 300 K).

Let's denote the wavelength at 1200 K as λ1 and the wavelength at 300 K as λ2. We have:

λ1 = h / sqrt(2mk*1200)

and we want to find:

λ2 = h / sqrt(2mk*300)

Dividing these two equations gives:

λ2 / λ1 = sqrt(1200 / 300)

Therefore:

λ2 = λ1 * sqrt(4) = 2λ1

So, the de-Broglie wavelength of the neutron at 27 °C is twice its wavelength at 927 °C.