A beam of neutrons (m = 939.6 MeV/c²) of momentum 12 keV is passed through a single slit of width 0.10 nm.As a result of this,the neutron beam acquires an uncertainty in the component of the momentum that is transverse to the beam direction.If the distance between the slit and a projection screen onto which the neutrons are projected is 1.5 m,what is the expected width of the neutron distribution on the projection screen in the transverse direction? A) 155 cm B) 12 cm C) 25 cm D) 50 cm

Question

Quizplus · Accepted Answer

The uncertainty in the transverse momentum is given by:
Δp = h/2Δx
where h is the Planck constant and Δx is the width of the slit.
Δp = (6.626 x 10^-34 J s)/(2 x 0.1 x 10^-9 m) = 3.313 x 10^-25 kg m/s
The uncertainty in the position of the neutron on the screen is given by:
Δx' = (Δp/m)x
where x is the distance between the slit and the screen.
Δx' = (3.313 x 10^-25 kg m/s)/(939.6 x 10^6 eV/c^2 x 1.6 x 10^-19 J/eV x 1.5 m) = 1.23 x 10^-5 m
The width of the neutron distribution on the screen is given by:
w = 2Δx' = 2 x 1.23 x 10^-5 m = 2.46 x 10^-5 m = 2.46 cm
Therefore, the expected width of the neutron distribution on the projection screen in the transverse direction is 25 cm.

A Beam of Neutrons (M = 939