The de - Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature `T ("kelvin")` and mass `m`, is

Options

A. `(h)/(sqrt(mkT))`
B. `(h)/(sqrt(3mkT))`
C. `(2h)/(sqrt(3mkT))`
D. `(2h)/(sqrt(mkT))`

Answer

B

Explanation

Thinking process de Broglie wavelengfth associated with a moving particle can be given as
`lambda=h/pr=(h)/(sqrt(2m(KE))`
At thermal equilibrium, temperature of neutron and heavey water will be same .
This commo n temp[erature is given as T.
Also, we know that, kinetic energy of a particle
`KE=(p^(2))/(2m)`
where,P = momentum of the particle
m=mass of the particle
Kinetic energy of the neutron is
`KE=(3)/(2)KT`
`therefore` de-Broglie wavelength of the neutron
`lambda=h/p=(h)/(sqrt(2m(KE))=(h)/(sqrt(2mxx(3)/(2)kT))=(h)/(sqrt(3mKT))`