Electrons with a speed of $2.1 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$ are directed towards a $1.0 - \mu \mathrm { m }$ wide slit. An electron detector is placed $1.0 \mathrm {~m}$ behind the slit. How wide is the central maximum of the electron diffraction pattern on the detector? $\left( h = 6.626 \times 10 ^ { - 34 } \mathrm {~J} \cdot \mathrm { s } , m _ { \text {electron } } = 9.11 \times 10 ^ { - 31 } \mathrm {~kg} \right)$
A) $690 \mu \mathrm { m }$
B) $1000 \mu \mathrm { m }$
C) $1400 \mu \mathrm { m }$
D) $350 \mu \mathrm { m }$

Question

Quizplus · Accepted Answer

The width of the central maximum in a diffraction pattern can be calculated using the formula for single-slit diffraction, $w = 2L\lambda / d$, where $L$ is the distance to the screen, $\lambda$ is the wavelength of the electrons, and $d$ is the slit width. The de Broglie wavelength $\lambda$ of the electrons is given by $\lambda = h / p$, where $h$ is Planck's constant and $p$ is the momentum of the electron. The momentum $p$ can be calculated as $p = mv$, with $m$ being the mass of the electron and $v$ its velocity. Substituting the given values, we find $\lambda = 6.626 \times 10^{-34} \mathrm{Js} / (9.11 \times 10^{-31} \mathrm{kg} \cdot 2.1 \times 10^6 \mathrm{m/s})$, which gives a wavelength. Then, using the formula for $w$, with $L = 1.0 \mathrm{m}$ and $d = 1.0 \mu \mathrm{m}$, we can calculate the width of the central maximum, which results in approximately $690 \mu \mathrm{m}$.

Electrons with a Speed Of 2.1×106 m/s2.1 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }2.1×106 m/s

Electrons with a Speed Of $2.1 \times 10 ^ { 6 } \mathrm {~m} / \mathrm { s }$