Electrons emerge from an electron gun with a speed of $2.0 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ and then pass through a double-slit apparatus. Interference fringes with spacing of $2.7 \mathrm {~mm}$ are detected on a screen far from the double slit. What would the fringe spacing be if the electrons were replaced by neutrons with the same speed? $\left( h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} , m _ { \text {neutron } } = 1.675 \times \right.$ 10-27 $\mathrm { kg }$ )
A) $0.93 \mathrm {~nm}$
B) $4.9 \mathrm {~m}$
C) $1.2 \mathrm {~km}$
D) $1.5 \mu \mathrm { m }$

Question

Quizplus · Accepted Answer

The fringe spacing in a double-slit experiment is directly proportional to the de Broglie wavelength of the particles, which is given by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle. Since the momentum $p = mv$, where $m$ is the mass and $v$ is the velocity, and both electrons and neutrons have the same speed in this scenario, the wavelength $\lambda$ is inversely proportional to the mass. Neutrons are much more massive than electrons, so their wavelength, and thus the fringe spacing, will be smaller for neutrons. The calculation involves finding the ratio of the electron mass to the neutron mass and applying it to the given fringe spacing for electrons to find the new spacing for neutrons.

Electrons Emerge from an Electron Gun with a Speed Of 2.0